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2000 Munkres
Topology: Solutions > Chapter 3 Connectedness and Compactness


Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Munkres, Section 23 Connected Spaces

1 If is disconnected then there is a separation in the topology which is also separation in . Equivalently, if is connected then so is .

2 If is disconnected then there is a separation of the union and each set of the sequence lies within either or . Suppose that , then by induction each and is empty. Contradiction.

3 If there is a separation of the union, then lies within or . Suppose, , then each and is empty. Contradiction.

4 There are no two disjoint nonempty open subsets.

5 In the discrete topology every subset is open, and, therefore, if a subset has more than one element, any pair of its elements can be separated. The converse does not hold: see Example 4.

6 If does not intersect the boundary, then and is a separation.

7 No, is a separation.

8 Argh! Suppose it is disconnected, so that there is a set which is both open and closed. For any sequence there is , and for any : there is . A sequence differs from by less than in each coordinate and it still have to be in . And similarly for and in . So, and must be preserved under bounded permutations (for example, in the uniform topology, i.e. the set of sequences converging to 0, would not work). At the same time, if is the set of bounded sequences, and is the set of unbounded sequences, then both seem to be open in the uniform topology. So, the answer is no, it is not connected.

9 Let . For any : is connected as a set homeomorphic to . And for all : is connected as well. Suppose and . Then is connected (the two sets have the point in common). Moreover, for every and : and intersect and . Therefore, by 3, is connected.

10 (a) It is homeomorphic to a finite product of connected spaces. (b) They all have a common point (). (c) This is the main part of the proof. Any point differs from at an infinite number of coordinates. Any its neighborhood has only finite number of coordinates restricted by some proper open subsets. A point that is equal to at these coordinates, and equal to at all others, belongs to the neighborhood and to . Therefore, the closure of is the whole space, and by Theorem 23.4 it is connected.

11 If were not connected, there would be a separation of and each must lie within one of the components of the separation. Therefore, each component of the separation is saturated, but by Corollary 22.3 there is a homeomorphism between and the quotient space, which implies that is disconnected as well. Contradiction.

12 It is sufficient to show for one set only, so we show that is connected. Suppose it is not, then let where and and are open in . . Using Lemma 23.1, no limiting points of can be in and no limiting points of can be in , so that is closed and is open in . But no limiting points of can lie in or . So that is closed in . Therefore, is both open and closed in . Contradiction.

Munkres, Section 24 Connected Subspaces of the Real Line

1 (a) The three sets in the question are all of different order types, but the question is whether you can construct a homeomorphism of their topologies regardless of the order itself. If we remove a point from each of two homeomorphic spaces they will remain homeomorphic. If one space is connected so is the other. Remove 1 from (0,1], it is still connected, but any point removed from (0,1) makes it disconnected. So that (0,1] is not homeomorphic to (0,1) (for similar reason [0,1) is not homeomorphic to (0,1) — we need it later, neither is [0,1]). The only two points that being removed keep [0,1] connected are 0 and 1, but since (0,1) is not homeomorphic to [0,1) or (0,1], we conclude that (0,1] is not homeomorphic to [0,1].

(b) , , , .

(c) Removing 0 makes disconnected but removing any point from leaves it connected.

2 Example 6 of §18 proves that is connected. Let be the continuous function (§21). For any : , therefore, by the Intermediate Value Theorem, there is such that or .

3 Suppose and . is connected (in all three cases). Let . It is continuous (§21) and , . By the Intermediate Value Theorem there is such that or . If or the problem is that we cannot take the values at both 0 and 1 and apply the theorem. In fact, does not have a fixed point if 1 is not in .

4 If there were an empty interval then would be a separation, therefore, for any there is . To show the least upper bound property suppose that is nonempty and bounded from above. Let be the set of all upper bounds of , it is nonempty. All we need to show is that has the least element. Let . Note that is an upper bound of iff for every : iff for every : iff is an upper bound of . Therefore, is the set of all upper bounds of as well. If has the greatest element then it is the least element of , so, suppose it does not. Then is open: if then, since there is no largest element in , there is such that . Since is connected, is not closed and there is a limit point of in . It is an upper bound of , and suppose there is such that . Then is a nonempty open set that contains but does not intersect . This contradicts the assumption that is a limit point of . So is the least element in and the least upper bound of .

5 (a) Yes, it is a linear continuum. For we have a point between them (consider two cases and ). If is bounded from above by then is bounded from above by and it has the greatest element . Then let . If then is the least upper bound, and if then is the least upper bound. (See also the next exercise.)

(b) The other way around it does not work: there is no element between, say, and . Also, it is not connected.

(c) It is. If then is between them. If is bounded from above by , then is bounded from above by and let . If then let and is the least upper bound. If then is the least upper bound.

(d) Note that in (c) we needed both points and to be in the space: the first one for the case and the second one for the case and . This suggests that the set in (d) is not a linear continuum. Indeed, is bounded from above but does not have the least upper bound.

6 Indeed, for any two elements there is an element between them (consider two cases and ). We show the greatest lower bound property. If is a nonempty set then has the least element and the greatest lower bound of is . By the way, need not be a linear continuum (it is not iff has an element that has a successor).

7 (a) “Order preserving” implies “injective”. Therefore, is bijective. Order preserving also implies that preserves the order topology.

(b) It is increasing from to (this part may require using the intermediate value theorem if it was not proved before), therefore, it is order preserving and surjective and, by (a), homeomorphism. Therefore, its inverse is continuous.

(c) is obviously increasing and surjective. In the order topology on would not be open, while it is open in the subspace topology. Therefore, would be a homeomorphism had the order topology, but in the subspace topology there are too many open sets in which makes not continuous: is open in the subspace topology (not in the order topology) but which is not open in .

8 (a) Let and be two points in the product. Let and be a point such that where is a path connecting with in . Since any open subbasis set in the product is an inverse projection of a proper open set for some , the preimage is just which is open in . (The proof here repeats partially the proof of Theorem 19.6, which can be used here instead.)

(b) No, the topologist’s sine curve is an example.

(c) Yes, the composition of two continuous functions is a continuous function.

(d) Take any two points. If they are in the same set in the collection, there is a path between them. If they are not in the same set in the collection then there is a path connecting the first point to a common point of all sets in the collection and another one connecting the common point to the second point, the joint path is still continuous and is a path connecting the point. I believe a more general statement is also true: if a path connected subspace has a common point with any set in a collection of path connected subspaces, then the union of the set with the union of the sets in the collection is path connected (the proof must be similar — the only difference should be that the constructed path may consist of three parts).

9 For any point there is an uncountable number of lines passing through it which do not intersect . For any two points there is a pair of lines that do intersect each other but do not intersect the set . So, both points are connected to the point of intersection of the lines, and, therefore, are connected.

10 If let such that there is a path between and . Using 8(d), is path connected. If then there is a basis neighborhood of , is a ball, and is path connected with any point in (just take the closed line interval connecting the points). Therefore, and is open. For similar reason, if then there is a neighborhood of contained in . If some point in were path connected to then so would be , but . Therefore, is open. Then is closed in . is not empty as , so if is not empty as well, then there is a separation of . Therefore, is path connected.

11 is connected in and its interior is connected but its boundary is not. Two balls in touching at a common boundary point is a connected set (as the union of two connected sets with a common point) and its boundary is connected (for the same reason), but its interior is not connected (as the union of two disjoint open sets). At the same time has connected interior and boundary but is not connected.

12 (a) If is a bijection preserving the order, then is a bijection preserving the order, similar for . Vice versa, if there are two order preserving bijections from and to and then the joint function is the order preserving bijection needed.

(b) and has the order type of implies, by applying (a) two times, that has the order type of . Vice versa, we construct bijective order preserving functions and the resulting function is an order preserving surjective function onto .

(c) Transfinite induction: we show that if it holds for any then it holds for . If has an immediate predecessor then is either empty or has the order type of , and has it as well. By (a), the property holds for . If has no immediate predecessor, then the set of all elements lower than has it as the supremum (for any there is ) and we can construct a sequence of points converging to . This is not a trivial fact, I think (and I don’t remember it was covered anywhere before). Indeed, if we consider , for example, then is a limit point of but there is no sequence converging to it (any sequence is a countable subset and has an upper bound in ). However, in the case of we prove it as follows: is countable; let us enumerate all its elements with positive integer indexes: ; then let be such that , let be the minimum index greater than such that , let be the minimum index such that and , etc. (we can always do this as there is no largest element in , therefore, there is infinite number of elements greater than any ); note that for every , and for all ; if then for some and for all . Now, once we have the sequence converging to , by using (b), has the order type of . By the principle of transfinite induction, this holds for any .

(d) We show that any point is path connected to where is the immediate successor of . Indeed, for is path connected to ( is homeomorphic to ) and for any point , , is path connected to which is path connected to (by (a) and (c), has the same order type as and the order topology on a convex subset is the same as the subspace topology). Therefore, is path connected as the union of path connected subsets having a common point.

(e) An open interval in has the order type of an open interval in . Indeed, by (c), has the order type of . Since for convex subsets the order topology is the same as the subspace topology, has the order type of . Therefore, for any there is an open neighborhood with the order type of . Moreover, there is a bijective order preserving function from one set onto the other, and by 7(a), it is a homeomorphism.

(f) If could be embedded into then would be homeomorphic to a subspace of which has a countable basis for the topology. Therefore, there must be a countable basis of the topology of . We show that there is no. For any : is open and contains but does not contains for . Therefore, there is a basis element such that it contains but does not contain for . Call it . Note that for : . Since is uncountable, there must be uncountable number of distinct sets in the basis.

Munkres, Section 25* Components and Local Connectedness

1 Suppose is connected and . is both open and closed in and nonempty. Therefore, using §23 criterion, for any and . So, each component of is a singleton. Hence, path components are also singletons. Now, the image of a connected space under a continuous mapping must be connected. Therefore, only constant functions are continuous functions from to . Compare to §18, exercise 7(b).

2 (a) Since is path connected, in the product topology is path connected and connected (§24, exercise 8(a)), therefore, there is one component = one path component, i.e. the whole space.

(b) preserves the metric, and therefore, by exercise 2 of §21, it is a homeomorphism. Hence, and are in the same component iff and are in the same component. We show that it is iff is bounded.

First, we show that is path connected to any other point where . Indeed, is a path. To see this, recall that a function to the subspace is continuous iff it is continuous as a function in to the space, and that the topology in a subspace is generated by the same metric as the topology of the space. Moreover, iff iff . Therefore, the preimage is open.

Now, two things follow. First, the space is locally path connected. Therefore, two points are connected iff they are path connected. Second, and are path connected iff is bounded. Indeed, according to the exercise 8 of §23, the sets of bounded and unbounded sequences is a separation, therefore, any point connected to (which is bounded) must be bounded. The other direction: if a sequence is bounded, then the argument similar to one in the previous paragraph shows that there is a path.

(c) If then is a path from to . Indeed, there is a finite number of nonzero coordinates in , and, therefore, the preimage of any open basis set of the box topology is a finite intersection of intervals open in . On the other hand, if then there is an infinite sequence of indexes such that and the homeomorphism if or otherwise (it is a homeomorphism by exercise 8 of §19), maps to and to an unbounded sequence. But the sets of bounded and unbounded sequences are open in the box topology (they are open even in the coarser uniform topology) and form a separation. Therefore, and must be in different components. BTW, the result also implies that in the box topology is not locally path connected, still components and path components are the same.

3 Any open interval in the ordered square is a linear continuum, therefore, it is connected. At the same time any open set that contains for contains some point , , and the argument similar to Example 6 of §24 shows that it is not path connected. Similar for points , . The space is locally path connected at all other points. (Another way to prove that it is not locally path connected is as follows: if it were locally path connected then its components would be the same as its path components, i.e. it would be path connected which contradicts Example 6.) The path components are the vertical segments. By the way, in the subspace topology as a subspace of in the dictionary topology (which is strictly finer than the order topology of the ordered square, remember exercise 10 of §16?) it is locally path connected but still not path connected (it is not even connected in this topology).

4 (Exercise 10 of the previous section is a particular case of this general result.) In an open subset of a locally path connected space each path component and their unions are open in and, therefore, open and close in . Since is connected, there is only one path component. Another way to show this is as follows. We prove that an open subspace of a locally path connected space is locally path connected (this implies that the components and the path components of the subspace are the same). Indeed, if is an open subset of a locally path connected space , and is an neighborhood of in then is open in as well (since is open) and contains a path connected neighborhood of which is also open in .

5 (a) Every point is (path) connected to , hence, the space is (path) connected. For the same reason it is locally (path) connected at . But for any other point every small enough neighborhood consists of disjoint line segments and is not (path) connected.

(b) The idea of the example in (a) is that every path connecting two points on a pair of line segments goes through a point, and, therefore, at any other point a small enough neighborhood that does not contain that point but still intersects an infinite number of line segments is not connected. We can extend the example to require the path for some points to go through two point. Let , then is (path) connected (as the union of the path connected line segment and the collection of other path connected line segments, each of which intersects ) but not locally connected at any point (now it is not locally connected at or and for all other points the argument remains as before).

6 Here we show as the hint suggested that a component of an open subset of is open. Let be open in and be its component. For any point there is a connected subspace and an open neighborhood such that . Since is a connected subset of , . Therefore, is the union of for all and is open. Another way to prove this is to show that the space is locally connected at every point. This way might be better considering the next exercise (in the proof we should somehow use the fact that the space is weakly locally connected at every point, or at least at every point in some neighrborhood of , to prove that it is locally connected at ). Let be a neighborhood of . There is a connected subspace such that it contains a neighborhood of . Suppose there is a neighborhood of such that the space is weakly locally connected at every point in the neighborhood. The intersection of these two neighborhoods of is a neighborhood of such that it is contained in and the space is weakly locally connected at every point in the neighborhood. If is a component of containing then every point in is contained in a connected subspace that a) has to be contained in and b) contains a neighborhood of the point. Therefore, as before, we conclude that is open in , and, therefore, in , contains and is connected. I.e. the space is locally connected at . Note that for this to be true we need the space to be weakly locally connected not only at the point but at any point in some neighborhood of . The next exercise shows that weak local connectedness at the point only does not imply the local connectedness at the point.

7 It is not locally connected at : for any bounded open neighborhood of its intersection with is open and bounded, and let be its and be the minimum index such that . Then may contain some points for but it does not contain . At the same time it does contain some points of and, therefore, some points of the line segments for sufficiently large . Those parts of the set are disconnected and the neighborhood can be separated. (A more formal prove would require a more formal statement of the problem and some technical work.) Now, for any neighborhood of we may find an open ball such that its intersection with the set is contained in the neighborhood, then find some such that and their line segments are in the ball — this will be the connected subspace, which does contain another neighborhood of . The reason the previous exercise argument does not work here is that the set is not (weakly) locally connected at some points in any neighborhood of , and the components can’t be represented as the union of open sets.

8 Let be a component of an open set . Since is saturated, it is enough to show that it is open. Since is locally connected, every component of every its open subset is open, therefore, it is enough to show that if then where is the component of in (which is open). is connected, therefore, is connected. It follows that and .

9 Following the hint. Let . We show that is connected iff is connected iff is connected. Indeed, this follows from the fact that and are homeomorphisms. Therefore, if is the component of then each of and is the component of containing and .

10 (a) It is reflexive and symmetric. To show transitivity suppose that but . There is a separation of points and , and is in one of these open sets, i.e. it is either separated from or . Contradiction.

(b) If for two points in one component there is a separation of these points, then the component is not connected. On the other hand, what can make a component to be a proper subset of a quasi-component? Or, equivalently, when a quasi-component may contain two or more components? Obviously, it cannot be the case if the components are open, as otherwise they would be open and closed and form a separation for points in different components. Therefore, if the space is locally connected, then the components are open, and quasi-components are the same as components.

(c) is not (locally) (path) connected. For any two points with different -coordinates consider a point such that and . Then open sets and form a separation of the points. Now, each vertical line segment is path connected, therefore, it is a path component, component and quasi-component. The remaining two points are disconnected, therefore, there are two more (path) components. The only question here is whether there is one or two quasi-components. We show that there is only one quasi-component with both points in it. The idea is that any separation of the space must “draw a line” between a pair of vertical lines without intersecting any of them, and therefore, the two points must be in one set in the separation. Suppose there is a separation such that and . For each open set and take a small open ball-neighborhood of the corresponding point contained in the set. The balls intersect some common vertical line, therefore, and intersect that line and form its separation. This contradicts the fact that the line is (path) connected.

is connected ( is path connected and adding a limit point keeps it connected), so that there is one component = quasi-component. But there are two path-components: and all other points.

is not locally (path) connected. Its path components are the line segments (so there are infinitely many of them). We show that it is connected and, hence, has one (quasi-)component. Suppose there is a separation . Both are open, therefore, they cannot intersect the same line segment. Moreover, they do not contain limit points of each other. Take any point and suppose it is in . It is a limit point for the set of line segments . Therefore, starting from some all lines are in . Therefore all points are in . Therefore, all line segments are in . This implies all their limit points in . By continuing this argument we conclude that all points are in , and must be empty. Contradiction.

Munkres, Section 26 Compact Spaces

1 (a) If is compact then is compact.

(b) Suppose one is finer than the other. Then the identity mapping from the finer one to the coarser one is a continuous and bijective function that maps a compact space to a Hausdorff space. Therefore, it is a homeomorphism and the topologies are the same.

2 (a) Any open set in a collection that covers a set covers all but finitely many points of the set. Therefore, a finite subcollection covers it all.

(b) No. Let .

3 For each subspace in the collection chose a finite subcovering that covers it and then take the finite union of all these finite subcoverings.

4 It is a Hausdorff, therefore, compact implies closed. If it were not bounded, then for any ball there would be a point outside it, and while the union of all these balls does cover the whole space, there is no finite subcovering for the subspace. See also the proof of Theorem 27.3. Now we must find an example of a space where there is a closed bounded set such that it is not compact. Boundness is not a topological property. Therefore, we can take any non-compact metric space and make it bounded by taking the standard bounded metric. Or, for example, take an infinite space in the discrete topology (with the discrete metric: for ).

5 We know from Lemma 26.4 that for each there are open sets and such that . covers and there is a finite subcovering . The finite intersection of corresponding sets is open, contains and does not intersect . (The proof is also given as a part of the proof of Theorem 32.3.)

6 Let be closed in . Then is compact. Then is compact and, since is Hausdorff, is closed.

7 Let be closed in and suppose . Suppose then and, since is closed,  is open and contains a slice . By the tube lemma there is also a tube about the slice that does not intersect . Its projection is an open neighborhood of that does not intersect .

8 Suppose the graph is closed and is open in . If then is empty and open. Now suppose that is a neighborhood of for some . Then is open in the product and is closed and so is its intersection with the graph. Using exercise 7, its image under the projection on is closed as well, but the complement of the image is the preimage , and it is open. Therefore, is continuous.

Now, suppose that is continuous. We take a point outside of the graph and show that there is its open neighborhood that does not intersect the graph. For points and take two neighborhoods and such that they do not intersect (Hausdorff property of ) and then take open neighborhood of such that ( is continuous, hence, is open and contains ). Then is an open neighborhood of such that it does not intersect the graph.

9 Suppose and are arbitrary spaces and is compact, then if open in the product contains then it contains for some neighborhood of and of . Indeed, each point has a basis neighborhood contained in . The union of the basis neighborhoods covers the set which is homeomorphic to the compact set and, therefore, there is a finite subcover. The finite intersection of ’s and the finite union of ’s are open and their product covers and is contained in .

Now, for each consider the basis element that contains and is contained within . The union covers and, since it is compact, there is a finite subcover. The finite union of ’s and the finite intersection of ’s are the open sets in and we are looking for.

10 (a) For each let be such that . Now, since is continuous, there is an open neighborhood of such that for each : . Since is monotone increasing, for all greater this still holds. Moreover, all ’s cover and, since it is compact, there is a finite subcover . Let be the maximum of . Then for each there is some nearby such that and for all : .

(b) The example of exercise 9 of §21 can be restricted to the compact domain . is a monotone increasing sequence converging to the constant function , but not uniformly (for any : ).

11 Just follow the hint. is closed. Therefore, both sets and are closed in . Since is compact, they are compact as well, and, by exercise 5, since is also Hausdorff, there are open and disjoint and containing them. are closed, nonempty (otherwise, since and are disjoint and both contain a point of , there would be a separation of , which is connected) and nested (here we use the fact that the initial sequence is nested, otherwise the result would not hold), therefore, since is compact, the intersection is nonempty. That implies that there is a point in that is not in and, therefore, not in . Contradiction.

12 In this exercise we are asked to prove that is compact given that is continuous, surjective and closed, is compact and the preimages of all singletons are compact. We prove a little more: if is continuous and closed and the preimage of a point is compact then the preimage of any compact subset is compact. If, for example, we knew that is compact and is Hausdorff then the requirement that “ is continuous” would be enough to guarantee that the preimage of any compact set is compact. On the other hand, if is compact and is just a -space, then what follows from what we are about to prove is that “ is continuous and closed” is enough to guarantee the same thing.

Suppose is continuous. When we can guarantee that the preimage of any compact set is compact? A constant function is an example that shows that we may need to require that the preimage of any point is compact. So, suppose it is. Let be a compact subset of and . We know nothing about : it need not even be closed. Let be an open covering of . covers . We know that for each there is a finite subcovering of . If we only could group all ’s in a finite number of groups where contains and is such that covers not only but as well, then we are done. Take any collection of neighborhoods over all . Since is compact, there is a finite subcollection that covers . Therefore, we just need to find a neighborhood for each such that covers as well. In other words, what we need for is the property that if an open set contains then there is an open neighborhood of such that . Now, this property is equivalent to the requirement that any is an interior point of where is open and contains . The requirement that is closed is obviously sufficient for this. (How did we use the continuity of ?)

13 (a) Using exercises 4 and 7(c) of §22, is closed and there are two disjoint open sets and containing given and , respectively. We show there exists some neighborhood of such that . Indeed, for each there is a neighborhood of such that (using the fact that is open and the multiplication is a homeomorphism, exercise 4 of §22), and their union is the neighborhood of such that . Now, cover and there is a finite subcovering. The corresponding finite intersection is the neighborhood of disjoint from .

(b) If is closed in and then which is closed by (a). And there is a neighborhood of such that it does not intersect . Then, is open (exercise 5(c) of §22), contains and does not intersect . Therefore, is closed.

(c) is continuous and closed (by (b)), moreover, is compact as the multiplication is a homeomorphism. Therefore, by exercise 12, as the preimage of is compact.

Munkres, Section 27 Compact Subspaces of the Real Line

1 Let be nonempty and bounded above by . Let . is closed and compact in the order topology (which is the same at the subspace topology, since the closed interval is convex). Consider the collection of all intervals such that and is an upper bound of . is nonempty as . None of those intervals are empty, and they are all closed in . Moreover, the intersection of any finite number of intervals in is also an interval in and, therefore, is a nonempty set. Given that is compact, the intersection of all intervals in has a point . Also, we have for all (in or below) and for all upper bounds of (in or above). There cannot be two such points , as in this case both have to upper bounds of but then . Therefore, there is a unique point in the intersection which is the least upper bound of .

2 (a) iff for any : iff .

(b) is compact, is continuous on and, therefore, continuous in both variables, hence, for a given : reaches the minimum on .

(c) iff for some iff .

(d) For let be such that . Balls cover , there is a finite subcovering , let be the minimum of corresponding ’s. For : .

(e) The idea is that there is no finite subcovering and as the minimum of ’s must be zero. For example, for has no -neighborhood in .

3 (a) The topology is strictly finer that the standard topology on which is compact and Hausdorff, therefore, it is not compact. Cool, huh? Another way to show this directly is as follows: has no finite subcovering. Yet, another way is just to say that is a closed subspace of in which is not compact (the set is infinite and all points are isolated), therefore, is not compact.

(b) Both topologies induce the same topology on and (it is clear for the first one, while for the second a subset is open in the subtopology of iff it equals the intersection of the subspace with an open set in minus some points in iff it equals the intersection of the subspace with minus those points in iff it is open in the subtopology of ), therefore, both are connected in either topology. Their closures are connected and share the common point , therefore, their union — the whole space — is connected.

(c) Suppose there is a path from 0 to 1 in . It is a continuous function from a compact connected space, therefore, the image is compact and connected in . Hence, it must also be connected as a subspace of (as it has a coarser topology), i.e. the image is convex and contains the whole interval . This implies that is a closed subspace of the compact image, i.e. it is compact in . Contradiction.

4 Given : is a continuous function in that maps a connected space into , therefore, the image is a connected subspace of that includes . This implies that it is either () or uncountable.

5 For any closed subset of the space (which is a compact subspace) and a point not in it there are disjoint neighborhoods, therefore, there is a neighborhood of the point such that its closure does not intersect the subset. Now we may proceed as in the proof of Theorem 27.7. We take any open set and show that it has a point not in the union. For , , there is a point in an open set that is not in ( has no interior points). There is a neighborhood of such that its closure does not intersect . Let . We have a nested sequence of closed sets , its intersection is nonempty as the space is compact, and any point in the intersection belongs to but does not belong to .

6 The set consists of closed intervals of length and the distance between them at least . Indeed, this is true for . At every step a closed interval of is divided into three parts with two parts in such that the distance between them is .

(a) For any two points there is such that they cannot lie in the same closed interval of , and, therefore, there is a point between them not in the Cantor set, and the set can be separated by and .

(b) It’s a closed subspace of a compact space.

(c) We have already shown by induction that it is the union of closed intervals each of length . By the same argument endpoints of the intervals are never excluded (only interior points are excluded).

(d) Consider a sequence of closed intervals containing a given point of , for each one choose an endpoint which does not equal the point, and all the endpoints are in , by (c), and the sequence converges to the point. Therefore, it is a limit point of the set of all other points in .

(e) It is nonempty (by (c), for example, or by the fact that it is the intersection of a nested sequence of closed intervals), compact and Hausdorff with no isolated points. By Theorem 27.7, it is uncountable.

Munkres, Section 28 Limit Point Compactness

1 Interesting. has no limit point (if has a coordinate in then a small enough ball does not contain any other point in , otherwise the distance to any point in is either 0 or 1).

2 does not have a limit point (if there were a limit point it would be a limit point in the standard topology as well, i.e. 1, but in the given topology is open).

3 (a) No. If is compact the image is compact, so if we believe the statement is wrong, we need to look for which is limit point compact but not compact. Consider in the Example 1. The projection on the first coordinate is continuous (it can also be thought of as the quotient space obtained by identifying all points in ) but maps the limit point compact space to the not limit point compact set .

(b) Yes. An infinite subset of has a limit point in which is a limit point of as well, i.e. it is in .

(c) No. Once again for the counterexample we need a limit point compact space which is not compact. Now the Example 2 works better: . Note that not only Hausdorff but also compact (and, hence, limit point compact).

4 Limit point compactness implies countable compactness for a -space. Let be a limit point compact -space and be a countable open covering of such that there is no finite subcovering of . Let . Note that for every , does not cover , but for every there is minimal such that . Let . For each let . This defines an infinite subset of that must have a limit point . But then the neighborhood of contains only finite number of elements in the sequence. And for each of them different from we can find a neighborhood of that does not contain it (here we use the -property — consider Example 1 to see why it does not work in general). The finite intersection of all these neighborhoods with is a neighborhood of that does not contain any point of the sequence different from . This contradicts the fact that is a limit point of the sequence.

Countable compactness implies limit point compactness. Suppose is a countably compact space and is an infinite subset of . There is a countably infinite subset and every limit point of is a limit point of as well. If no point in is a limit point of then every point in has a neighborhood that does not contain any other points in , and the countable collection of such neighborhoods covers . Since each set in the collection covers one point of only and is infinite, there is no finite subcollection covering . Therefore, is not closed (a closed subset of a countably compact space is countably compact: add the complement of the subset to any its covering) and there is a limit point of .

5 If is countably compact, then since has no finite subcovering of , it does not cover . The other direction: suppose that there is a countable covering that has no finite subcovering, then is a nested sequence of nonempty closed sets and it has a nonempty intersection. This contradicts the assumption that is a covering of . Note that the second part of the proof works for countable coverings only. For uncountable coverings there is no general way to order them in a sequence so that the sequence will contain all the sets in the collection and the complements of the partial unions of the sequence will generate a nested sequence of closed sets. Instead, we use the finite intersection property.

6 is injective, we show it is surjective. If , then since is continuous and is compact and closed, there is an -neighborhood of that does not intersect . Now let and . Then for all and for : . Therefore, no subsequence converges (if it did converge it would be within -ball of some point starting from some index). Now, is a compact metric space, therefore, it is sequentially compact. But we just constructed a sequence such that no its subsequence converges. Contradiction.

7 (a) and (b) Step 1. If is a shrinking map then there is at most one fixed point. Suppose there are two fixed points and . Then .

Step 2. We construct a closed compact set that will be proved to have a fixed point. Let and . is a continuous map from a compact space to Hausdorff space, therefore, it is a closed map. Then, by induction, every set in the sequence is closed. Moreover, since the image of a subset lies within the image of a superset, the sequence is a nested sequence of nonempty closed sets that has a nonempty closed and compact intersection .

Step 3. We show that . Indeed, suppose that for : . Hence, there is such that . This implies that which contradicts the fact that .

Step 4. We show that has only one point, this implies that the point is the fixed point. If is a contraction we obtain the result immediately. Indeed, the diameter of decreases exponentially and there cannot be more than 1 point in the intersection. More generally, we want to show that any is the image of some point in . For each : , i.e. there exists such that . A compact metric space is sequentially compact, hence, there is a subsequence of that converges to some point . Any neighborhood of contains infinitely many members of the sequence, therefore, it is in the closure of every , but ’s are closed, therefore, . Moreover, since for all and is continuous, . We conclude that , i.e. . The distance on is a continuous function from the compact product to the ordered set . Therefore, there is a pair of points with the maximum distance between them. Let where . Then if : . This contradiction implies that there is only one point in , the fixed point of .

(c) is strictly increasing from 0 to 1/2. implies it is a shrinking map (the unique fixed point is 0). implies it is not a contraction.

(d) is strictly increasing, moreover for all . So, there is no fixed point. but can be made as close to 1 as desired when . Therefore, it is a shrinking map which is not contraction and has no fixed point.

Munkres, Section 29 Local Compactness

1 are not compact as we may take a sequence converging to an irrational number (in ) and no subsequence converges to a point in (sequential compactness is equivalent to compactness for metric spaces). Suppose some compact (and, therefore, closed) subset of contains an open subset of . Then it contains an interval . The interval is closed in and, therefore, compact. Contradiction. Therefore, there are no compact subsets of that contain any open subset. Hence, is not locally compact.

2 (a) The projection is an open continuous map, therefore, we may use the next exercise to argue that all are locally compact. A compact subspace of the product containing an open set has all but finitely many projections equal to the whole corresponding space, since the projection is continuous, these spaces must be compact.

(b) Assuming the Tychonoff lemma all we need to prove is that the product of two locally compact spaces is locally compact. For any find the corresponding compact subsets and neighborhoods in both spaces and take their products.

3 What we need from the map is that it preserves both compactness and openness. The continuity guarantees that if is contained in a compact subspace then its image is contained in a compact subspace as well. The openness of the map guarantees that a neighborhood of within the compact subspace maps into a neighborhood of the image within the compact subspace containing it. If a continuous function is not open, in general, we cannot guarantee the second property. To find a counterexample we need to consider some space which is locally compact but not compact (otherwise the continuity will be enough). Moreover, it must be mapped onto a space that is not locally compact. By now I know three types of spaces which are not locally compact: (infinite products), (infinite totally disconnected) and (strictly finer than a compact Hausdorff topology). I think the second one should work. So we should take a locally compact but not compact space, for example, a locally compact unbounded subset of and construct a continuous function that would map it onto . Let and . Let . Then is a continuous (the preimage of any set is the union of open intervals) non-open (the image of a bounded open interval is not open in the range) map defined on a locally compact set whose image is exactly which is not locally compact. Cool!

4 Suppose where is open and is compact, then there are two balls centered at such that . But then as a closed subset of compact set must be compact, but it is not (similar to Exercise 1 of the previous section).

5 Let , then it is bijective and is open in iff is open in or is compact iff is open in or is compact.

6 The circle without a point is homeomorphic to the real line. Now using two facts, first, that the one-point compactification is unique up to a homeomorphism, and, second, that the compactification of the punctured circle is the whole circle (in fact, showing this is quite similar to the next exercise), we get the result.

7 : if we show that the latter space is compact and Hausdorff we are done. It is Hausdorff, and compact as any covering contains an open set containing , which in its order contains an interval . But the rest of the space is compact (as the ordering satisfies the least upper bound property).

8 is homeomorphic to the set in the discrete topology, which is equivalent to the topology inherited from the standard topology of the real line. is a compact and Hausdorff space, therefore, it is a one-point compactification of the subspace .

9 I believe the quotient map from onto is open, therefore, is locally compact (exercise 3). Yes, right: see the supplementary exercises of the previous chapter.

10 Let where is open and is compact. and is open. is closed in which is compact (and closed), therefore, is closed in and compact. We can separate and by neighborhoods and . It follows that is an open neighborhood of in such that is a closed (and compact) subset of .

11* (a) is not locally compact. Example 7 of §22. Let be a quotient map on that identifies all positive integers to a point . Let (it is not locally compact but Hausdorff). Then is not a quotient map. The open set in the example has the property that for any and there is a large enough such that is not in . In other words, there is no saturated neighborhood of, let say, point in for which there exists neighborhood of in such that there product is contained in . Any such neighborhood being saturated would contain all points and some their local basis neighborhoods . Any would contain an interval . Then for a large enough such that there would be a point in not in . Note that is a neighborhood such that it contains for any but does not contain any tube about it. This would not be possible if were compact.

is not Hausdorff. Exercise 6 of §22 is an example of a quotient map from a non-locally compact Hausdorff set onto a compact non-Hausdorff space such that the product of it with itself is not a quotient map. What I want is similar to the previous example with the only difference that is locally compact but not Hausdorff. Let us take a one point compactification of (in fact, the one point compactification was defined for locally compact Hausdorff spaces only, is not locally compact, but it seems that the Hausdorff property is enough to construct the one-point compactification, and local compactness is needed for the Hausdorff property of the constructed space only). Therefore, the constructed space is (locally) compact but not Hausdorff: the open sets containing the new point are those having bounded and closed in complement, such sets have no interior points in . Now, if we take the same example, then is still saturated and open, and its image is not open. The difference between this case and the previous is that here we made the space compact at the cost of being Hausdorff, hence, we still cannot guarantee that any neighborhood would contain a compact closure of another neighborhood — something that we need for the proof.

Now the proof. is continuous (see §18). Let be open. We show that is open. Let . Since is open and is locally compact there are neighborhoods of in and of in such that . If we knew that is saturated (or, alternatively, if we knew that is an open map) we would immediately conclude that has a neighborhood contained in . Since we don’t know that, we must proceed in finding a saturated neighborhood of such that . Since is saturated, we have . In the first example, suppose we take point (1,0), then find a “rectangle neighborhood” about it, then the image of this rectangle has point , therefore, the preimage of the image has all points (the preimage of the image of the rectangle neighborhood contains the original rectangle about (1,0) plus all other points with vertical intervals). Now we use the fact that is compact. For each point in we take a neighborhood such that its product with is still in . We call the union of all this neighborhoods . This is exactly what we could not do in the first example: for any there was a large enough point such that we could not find the desired neighborhood containing it due to the fact that no were compact. Now we proceed this way constructing . Note that for each , and . Let . Then is open. If we show that is saturated, we are done. Indeed, , hence is saturated.

(b) This is the easier part. Using (a), we conclude that the composition of two quotient maps is a quotient map given that both and are locally compact and Hausdorff.

Supplementary Exercises*: Nets

1 Given two elements the one that is greater than both is: (a) the greater one (any pair of elements are comparable), (b), (d) the union of a pair of subsets, (c) the intersection of a pair of subsets.

2 Take two elements in , there is an element greater than both ( is directed), and an element greater than ( is cofinal).

3 If , which is ordered and, therefore, by 1, is a directed set, then the net is merely a sequence of points and the definition of convergence is as follows: iff for any neighborhood of there is such that implies .

4 Every neighborhood of contains a basis neighborhood . There are and such that and for all and . Since is a net, there is such that implies both and . Therefore, “starting” from all points are in .

5 Separate two points by two non-intersecting neighborhoods and . There are and such that and for and . There is such that implies both and which contradicts the fact that .

6 If there is a net in converging to then for any neighborhood of there is a point in , therefore, . Now, suppose . Using 1(c), consider the collection of all neighborhoods of partially ordered by the reverse inclusion (the “finer” is the set, the “greater” it is). Now, for each neighborhood take a point . Then, is a net of points of converging to . Indeed, given any neighborhood of and for : .

7 If is continuous, and is a neighborhood of , then is an open neighborhood of and for some , implies and .

The other direction. Let be open, , and . We need to show that there is an open neighborhood of contained in . Suppose there is no such neighborhood. Then . Using the previous exercise, there is a net of points of converging to . But in this case contains no points of which contradicts the assumption that the image of any net converging to converges to .

8 If then for any neighborhood of there is such that implies . Since is cofinal in J, there is such that . For all , , hence, .

9 If there is a subnet converging to then for any neighborhood of there is such that implies . Let . is cofinal in . Indeed, for any there is such that and such that is greater than both and . Then and . Now, the set of all points from such that contains , therefore, it is cofinal in .

The other direction. For any neighborhood of let be the cofinal subset of those such that . We need to define a directed set and such that implies , is cofinal in and for any neighborhood of there is such that implies . Consider two different neighborhoods and of and their intersection . Note that . However, we want to distinguish all indexes in from the same indexes in and . This suggests to take as the union of all pairs where and define . Moreover, for a given we want both: and . Therefore, we define the partial order on as follows: iff and . This is a partial order on (this can be easily checked even in general for the partial order defined this way on the product of two partially ordered sets). Now, for any pair of elements of , and : find , then such that (here we use the fact that all are cofinal), then we have . Therefore, is a directed set. Moreover, implies and is cofinal in . For any neighborhood of the set is not empty, and implies .

10 Suppose the space is compact. Take any net . For each let . Then, given that the index set is directed, the collection of sets satisfies the finite intersection property. Let be the intersection of the closures of these sets and be its neighborhood. Since lies in the closure of each , for each there is such that . Therefore, is an accumulation point of the net, and, using the previous exercise, there is a subnet converging to .

The other direction. Suppose that every net in has a convergent subnet. Let , be a collection of closed sets satisfying the finite intersection property. Consider the set of all finite subsets of : . It is a directed set with the partial order given by the reverse inclusion. Moreover, since each finite intersection has a point, for each we can take . This is a net in , and it must have an accumulation point . In particular, for every and neighborhood of there must be such that . Given that and we conclude that . Thus, lies within the closure of which is equal to . Therefore, is not empty.

11 Roughly speaking, if then there is a net in converging to , there is a subnet of points in converging to some ( is compact and closed as is Hausdorff), and the corresponding subnet of points in converges to a point in which must be ( is closed, the operations are continuous).

12 If we omit condition (2) in the definition of the directed set, then it is the same as to assume that there could be classes of different but “equivalent” indexes. Therefore, this does not restrict the previous definition, and all the examples in the Exercise 1 are still nets. The exercise 2 showed that a cofinal subset of a directed set is a directed set itself, and the proof remains valid as it does not depend on the property (2). The Exercise 3 shows that a sequence is a net, and since the new definition of a net enlarges the set of all nets, the fact still holds. In the exercise (4) we showed that a pointwise convergence of a finite product of nets implies the convergence of the product and, once again, the proof does not depend on the property (2). Exercise 5 tells us that in a Hausdorff space a net cannot converge to two different points at the same time, and the result is still valid as the proof depends on the fact for a given pair of elements there is another one greater than both whether the two are equivalent or not.

Now, we claim that for any net satisfying the new definition there is a subnet that is also a net according to the initial definition. Indeed, consider the new index set that consists of all equivalence classes of with the partial order given by the previous relation (i.e. iff ). It is straightforward to see that this new relation is well-defined and is a (strict) partial order on the set of all equivalence classes. Take the map by taking to be any element from . This map preserves the relation, and its image is cofinal in as for all . Let us call a subnet constructed this way (there may be many as the choice of is arbitrary).

If converges to some point , then converges to the same point. In fact, this holds more generally, namely, if a net converges to a point, then every its subnet converges to the same point. This was proved in Exercise 8, and the proof remains true for the new definition as well. Using this and the construction of a subnet satisfying the initial definition given above, we conclude that the theorem of Exercise 6 (a point is in the closure of a set iff there is a net in the set converging to the point) still holds.

For Exercise 7 we need a more general result. Suppose that for a net that satisfies the new definition every its subnet that satisfies the initial definition converges to . Then . [ Suppose not. Then there is a neighborhood of such that for every there is such that . For each define to be an element such that if such an element exists or any arbitrary element in otherwise. Then this defines a subnet of the type the same way as before, i.e. it satisfies the initial definition. For any : and . This contradicts the assumption. ] Given this result, we conclude that if is continuous and a net (in the new definition) converges to then every its subnet satisfying the initial definition converges to and its image converges to , therefore, the image of the net converges to as well. The reverse implication is immediate.

The lemma of Exercise 9 tells us that is an accumulation point of the net iff its a limit of a subnet. Note, that is an accumulation point of iff for any neighborhood of the set of all indexes such that is cofinal in iff the set of all indexes such that there is such that is cofinal in iff there is a subnet that has an accumulation point . This implies that there is a subnet converging to . The other direction: if there is a subnet converging to then we can leave the proof that it is an accumulation point the same as before.

For the Exercise 10 suppose, first, that every net has a convergent subnet. Then every net according to the initial definition has a convergent subnet that has a convergent subnet satisfying the initial definition. Therefore, as was proved before, the space is compact. Now, suppose that the space is compact. Then for a net there is a subnet satisfying the initial definition that has another subnet converging to a point.

Finally, the proof of the Exercise 11 does not depend on the definition of the net but rather on the previous results which we have already proved to hold under the new definition.

Permanent link to this article: http://dbfin.com/2011/03/2000-munkres-topology-solutions-chapter-3/


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  1. amir

    in section 24 question 8-a we have product of path-connected spaces.it means we have to work in product topology or it means the cartesian product of path-connected spaces?(i think that cartesian it means cartesian product)

    1. Vadim

      Yes, it means product topology, because if the product is not a topological space, then there is no notion of path-connectedness. Therefore, it is implied that the product is considered under some topology. If the topology is not specified, the product topology is typically assumed (Munkres specifically said somewhere before that the product topology is assumed unless otherwise is specified). On the other hand, it might be useful to check whether the product is necessarily path-connected in other topologies on the product as well, for example, under the box topology. This is not the case. For example, is not (path) connected under the box topology.

  2. AnthonnyAG

    Hi, in excercise 26.8 (<=) i don't see why if UxV' doesn't intersects the graph of f, makes it a closed set. Isn't it enough to consider only V and see that UxV is an open neighborhood of (x,f(x))? I did it that way but i fail to use the fact that Y is hausdorff.

    1. Vadim

      What I do here is I take a point outside of the graph (remember, ) and show that there is its open neighborhood that does not intersect the graph (I made it more clear in the solution). Hence, I show that the complement of the graph (which belongs to) is open and the graph itself is closed.

      Now, we need Hausdorff property in this proof because we need both open neightborhoods and that do not intersect. The former is needed to construct open set such that does not intersect , and, therefore, does not intersect the graph. The latter is needed for the neightborhood of itself.

      The question remains whether we need the Hausdorff property in general, i.e. not for this particular proof, but for any other proof as well. In other word, is it still true that the graph of a continuous function is closed even if we do not require to be Hausdorff (we still require it to be compact)? The answer seems to be no. Indeed, if has at least two points and is indiscreet, then is compact (but not Hausdorff), and any function is continuous (the preimage of any open set in under any function is either empty or the whole space ), while there are no graphs closed in (because open sets are all of the form where is open in , and, remember, there are at least two elements in , hence, the complement of any graph is not empty but not open).

      1. AnthonnyAG

        Thanks a lot for answering, now it’s clear for me. I was wondering if there is A difference making y not in Gf instead of y not equal to f(x). In this last case, is possible that y=f(x_0) for x_0 not equal to x?

        1. Vadim

          Ok, now it seems to me I understand the root of your questions. The graph is a set of points in space . It is not the same as the image of functoin. For example, let and . Then the graph is just a straight line . And is not on the graph iff . For example, is not on the graph even though there is another such that but it does not matter. Now, to show that the graph is closed we take a point outside of the graph, i.e. a point such that , and show that there is an open neighborhood in space containing and not intersecting the graph. But what does it mean that does not intersect the graph? It means exactly that for any and any we should have not in graph, i.e. . In other words we must have , and this is exactly how we construct (its image is in , and does not intersect ).

  3. Tbomas

    The proof for #23.4 is wrong. The finite complement topologo certainly has disjoint, open, nonempty subsets. Like the even and odd numbers, say.

    1. Vadim

      The finite complement topology is defined so that a subset is open iff its complement is finite. In particular, if the space itself is infinite, then every open set in it is infinite as well. But not vice versa, not every infinite subset is open. Like in your example, the subsets of even and odd numbers are not open subsets of the set of integer numbers in finite complement topology. They are open in discrete topology.

      Now, consider an infinite space with finite complement topology and two open sets and . The complement of is finite, and is infinite. Therefore, cannot be contained in , and, hence, and cannot be disjoint.

  4. Luc


    I think there is a mistake in section 13, question 7 with 4>2. I think 4 and 2 are not comparable.
    This guy thinks like me and explain better:

    Finally we have:
    J1 J2 J3 J4 J5
    J1 =
    J2 > = > nc >
    J3 < < = nc > = >
    J5 < < nc J1 => J2 finer than J1 and J2 nc J4 => J2 and J4 are not comparable


    1. Vadim

      His argument is wrong. And (-1,1)-K=(-1,0]U(1/2,1)U(1/3,1/2)U… is open in the upper limit topology. So, I think everything is fine here.

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