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2000 Munkres
Topology: Solutions > Chapter 2 Topological Spaces and Continuous Functions


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 12 Topological Spaces

No exercises.

Munkres, Section 13 Basis for a Topology

1 For every there is an open set such that , therefore, is open and , i.e. .

2 Let us enumerate the topologies by columns, i.e. we give numbers 1-3 for the first column from top to bottom, 4-6 for the second column, and 7-9 for the third column. Let “>” be the partial relation between any two topologies that indicates that the topology on the left side is finer (larger) than the topology on the right side. The finer (the larger) topology is the one with the “dust” not the “gravel”. Here we list all maximal ordered subsets of the set of topologies of Figure 12.1 ordered by “>”: 1<2<6<9, 1<2<7<8<9, 1<3<4<6<9, 1<3<7<8<9, 1<5<9.

3 The proof is very similar to Example 3 of §12: the empty set and are in the collection because their complements are and the empty set, the complement of any union of open sets is the intersection of the countable complements of the sets, so it is countable as well, finally, the complement of the finite intersection of open sets is the union of the countable complements, so it is countable. Now is the trivial topology if is finite, but if is infinite then it is not a topology, as, for example, we can partition into three disjoint sets such that the first two are infinite, and then and are open but their union is not.

4 (a) The empty set, , unions of open sets in the intersection and their finite intersections all lie in every topology in the collection, and therefore, in their intersection. The union does not have to be a topology. For example, the union of two topologies with bases consisting of half-open intervals, one (the lower limit topology) and the other (the upper limit topology), is not a topology, as, for example, the intersection of and is not in the union.

(b) The unique largest contained in all the topologies in the collection is obviously the intersection of the topologies. The smallest topology containing all the topologies in the collection is the one generated by the union as the subbasis.

(c) In Figure 12.1 these are topologies 4 and 5. Their intersection is the largest topology contained in both, it is . Their union is a subbasis for the smallest topology containing both, which is .

5 Every topology containing the collection must contain all unions of its sets, i.e. the topology generated by it. If the collection is a subbasis, then every topology containing it must contain all finite intersections of its sets, i.e. the basis generated by this subbasis.

6 For there is no open set in containing that lies in . For there is no open set in that lies in .

7 Let ">" be the "being finer" relation on the set of topologies (as in the Exercise 2). Then 4>2>1>3, and 1>5, but 3 and 5 are not comparable. The relations are easy to see using lemma 13.3 (we need to check whether it is true that every open set of one topology containing a point contains an open set from the other topology such that it contains the point). To see that 3 and 5 are not comparable consider point and two open sets and . Neither of the open sets contains an open set from the other topology that contains . (See the diagram on the right.)

8 (a) The standard topology is finer than the topology generated by . To see that they are equivalent consider any set open under the standard topology. Take any point . Since is open, and the set of all open intervals is a basis for the standard topology, there is an interval that contains and lies in . There are two rational points and such that . The interval contains , lies in and is a basis element in .

(b) The lower limit topology is finer than the topology generated by . Now, for point there is basis element in that contains and is a subset of .

Munkres, Section 14 The Order Topology

No exercises.

Munkres, Section 15 The Product Topology on XxY

No exercises.

Munkres, Section 16 The Subspace Topology

1 is open in iff and is open in iff and and is open in iff and is open in iff is open in .

2 It is finer but not necessarily strictly finer. It is finer, because if we change the topology to then all sets that were open are still open, and therefore their intersections with are still open in . It is not necessarily strictly finer as the new open sets in may not produce new open sets in the subspace topology. For example, a one-point subset will have the same subspace topology regardless of the topology on .

3 , and are open in , and and are open in .

4 Let be an open set, and . Then there exists such that . Since is open, there is a basis set in that contains . Since it is a basis set, is open in . Moreover, . Therefore, is open. Similarly for .

5 (a) Every basis set in is a basis set in .

(b) Yes. If is open in , , is open in , then is open in and, therefore, open in . Therefore, there exists a basis set in such that it is a subset of and it contains . Therefore, there are open sets and such that and . So, is open in and is open in .

6 A collection of sets such that is a basis for the standard topology on (see Exercise 8(a) of §13).

7 No, for example, is a convex subset of the real line.

8 A basis for is a collection of sets . A basis for is a collection of sets . In both cases the intersection of a basis set and the line is either or or or , and vice versa. Now not all combinations of these intervals are possible in all cases. If the line is vertical, then the respective topologies are and or (depending on the direction of the line). If it is horizontal, then in both cases the topology is either or . If it has some slope then the first topology may generate either ( and ) or ( and ), depending on the direction of the line. That is, it is either the lower limit or the upper limit topology. For the second topology, if the line is downward sloping, we have , , and , i.e. the discrete topology. Finally, if the line is upward sloping then the second topology generates a basis with either or , i.e. either or .


9 Every interval in the dictionary order topology on is a union of open sets in . Therefore, the later is finer than the former. A basis for the latter topology is a collection of sets which is an interval in the dictionary order topology. The topology is strictly finer then — see exercise 5(a): for example, is open in it, but not in .

10 The first two topologies are not comparable. Indeed, has no open neighborhood in the second topology, and has no open neighborhood in the first topology. The third topology is strictly finer than the first and the second one. Indeed, by 9, the third topology is generated by sets : every basis element of the first topology can be generated as the union of open sets in the third topology, and every basis set in the second topology can be generated as the union as well. The fact that the third topology is strictly finer follows from the fact that the first and the second topologies are not comparable.

Munkres, Section 17 Closed Sets and Limit Points

1 The proof is similar to the Theorem 17.1, just the other direction.

2 If is closed then we have: is closed in iff and is closed in iff and is closed in .

3 is closed.

4 is open in . is closed in .

5 is closed and contains , so it contains the closure of . It equals the closure if both endpoints are limiting, i.e. if is not empty and for every there are such that . This is equivalent to the requirement that has no immediate successor, and has no immediate predecessor (this requirement is sufficient as discussed above, and it is also necessary as otherwise if, for example, has immediate successor then is an open set containing that does not intersect ).

6 (a) is closed and contains , therefore, it contains the closure of .

(b) and (c) Every closed set containing the union contains each set, and therefore, its closure. The intersection of such closed sets (the left side) contains all closures of all sets (the right side). Now, the other direction.
For any finite number of sets : (in fact, this is the proof for both directions in the case of a finite number of sets).
For infinite number of sets: some collections of closed sets have non-closed unions so that the intersection over all closed is the intersection over smaller collection of sets. This suggests that we may have the right side as a strict subset of the left side if there is a collection of closed sets of such that their union is not closed and there is no closed set that is contained in the union and contains all the sets. For example, if we take and then .

7 The problem of the proof is that for different ’s the set may be different so that we cannot prove that lies in the closure of some particular (consider the example of the previous exercise).

8 (a) and (b) in both cases. Using 6(a), for every , and therefore, the left hand side is a subset of the right hand side. They do not equal in general, as there might be some common limiting or interior/limiting points which are not in the intersection and not limiting points of the intersection. For example, the intersection can be empty, while there is a common limiting point: and is an example.

(c) Using 6(a,b): . This implies .
Alternatively, assume that is in , i.e. every neighborhood of intersects but there is some neighborhood that does not intersect . Suppose there is some neighborhood that does not intersects , then the neighborhood does not intersects and it does not intersects , i.e. it does not intersects , contradiction. So, but . The other direction is immediate using 6(a).
The inclusion is strict and the same example and demonstrates the fact: .

9 iff for every basis set of , i.e. is open in and is open in , such that the intersection iff for every neighborhood of and every neighborhood of : iff and iff .

10 If then if there is such that then neighborhoods and are disjoint, otherwise and are disjoint.

11 If two points have equal -coordinates, i.e. the points are and , then let be a neighborhood of , and the disjoint neighborhoods of the points are given by product of and disjoint neighborhoods of and . Similar in the case they have equal -coordinates. Otherwise, the disjoint neighborhoods are products of disjoint neighborhoods of the projections of the points onto and .

12 The disjoint neighborhoods of two points in the subspace are the intersections of and two disjoint neighborhoods of the points in the space .

13 Cool! is closed in iff for every there are two open sets containing and containing such that for no point iff any pair of different points have disjoint neighborhoods.

14 For any point and any its neighborhood there is only finite number of points in the sequence that may be not in the neighborhood, so the set of “limits” is .

15 : is closed iff : is open iff : there is open such that and .

16 (a) and (b) The closure is the union of the set and its limiting points. So we describe only limiting points.

Hausdorff + + - + -
+ + + + -

17 In the lower limit topology and . In and . The difference for the set is exactly the one described in the solution for Exercise 8 of §13: set is open in the lower limit topology, but not open in (every open set in containing an irrational point has a rational point below it).

18 as every neighborhood of is for some and contains a point for sufficiently large . for similar reason. . .

19 (a) open open or open . Also, the two cases are disjoint, so that .

(b) open s.t. or .

(c) is open .

(d) No, is open, therefore, , but, for example, .


Set Bd Int
E or
F and or and

21 We use the following notation here: we define four operators as follows, , , , . So that, for example, and (19(a)). First, several statements:

  1. and are monotone: if then and . Indeed, every open set contained in is contained in as well, and every closed set containing contains as well.
  2. , , , but in general. Indeed, is a closed set containing . . is an open set contained in . .
  3. and is a partion of (the sets are disjoint). This follows from the definition of the boundary, 19(a) and the fact that .
  4. , , . Using 19(a) and properties 2 and 3, . Then, using this and property 2, , and .
  5. , . Indeed, and is open. and is closed. Both can be strict, as, for example, for .
  6. , and are, in general, not comparable by inclusion. Consider the following set: . Then and , and each of the following points belongs to one set only: , and .
  7. , . By properties 2 and 5, , but , therefore, . Now, using this and properties 2 and 4, .

The property 4 shows that even if we take as an additional operation we are not going to get more sets. The property 2 shows that in a sequence of operations of closure and complementation we can substitute any two operations of the same type by one operation of the same type. Therefore, we may construct only two sequences of sets: starting with two initial sets and we apply the sequence of operations . Now, properties 4 and 2 are especially useful for moving operators to the right (by switching it with another operator on the right: and ) and for canceling two 's in a row (). The first few members of the sequence starting with are , , , , , , , , and the second sequence starts with , , , , , , , . Note that, by 7, in each sequence the 8th member equals the 4th member, therefore, there can be at most 7 distinct sets in each sequence. Note also, that each set starting from the second one is open or closed. Let be neither open nor closed. This ensures that no other set in the sequence equals or . The following table splits the sets in the sequences in 4 different groups according to whether a set is a sequence of and starting at or , and whether it is closed or open.


The sets in different groups can easily be made different. For example, closed sets and open sets are different in the standard topology on the real line. Also, since interior is preserved, we can have two different points in and such that each one will be in every set in one column but not in any set in the other column. So, the challenge is to try to make all sets within one group different. For example, to make and different must have a limiting point such that it is an interior point of but not an interior point of . The set we used to illustrate property 6 is an example of how this can be done. But that example will not work as for that set : . We also need to add some isolated points to the set. Now, the same should work for the complement as well: to create an isolated point for we need a "punctured" subset in . Here is an example:

In both cases the next closure equals the 4th member of the sequence. So, there are maximum of 14 sets that can be obtained from a given set by taking closure and complement (and interior, as the latter can be expressed in terms of the former two) operations.

Munkres, Section 18 Continuous Functions

1 Let be open and be continuous in the definition. Then, implies and and s.t. implies . Therefore, is open.

2 No, for example, can be a constant function.

3 (a) Both mean that “any set open in is open in ”. (b) Follows from (a).

4 Let . Then is open in iff is open in iff is open in . Similarly for .

5 In both cases works as a homeomorphism.

6 For example, if or otherwise.

7 (a) If is open in and then and for some . Therefore, s.t. and . Thus, is open in .

(b) From to only constant functions: for any the inverse image has to be both open and closed, and there are only two sets in , namely the empty set and , such that they are both open and closed. Now, from to . is continuous iff for any and there exists such that . This can be proved exactly the same way as it was proved one direction in the text (Example 1), and the other direction in the Exercise 1. Now, what does this definition of continuity means... must be continuous from the right and non-decreasing.

8 (a) is open.

(b) for such that , and for such that , both sets are closed by (a), and restricted on them is continuous. Using the pasting theorem, is continuous.

9 (a) The pasting lemma applied several times.

(b) on , on are continuous, but on is not. The reason is that the inverse image of a closed interval is a union of the closed inverse images, but if the union is not finite then it may not be closed, as for in the example above.

(c) Let be a closed subset of . Then (this follows from the fact that ). Suppose . There is a neighborhood of such that it intersects only a finite number of sets in the collection: . For each : is closed in A_i and, therefore, closed in (as is closed). Moreover, . Hence, there is a neighborhood of such that . The intersection is a neighborhood of such that . We conclude that is closed.

10 The inverse image of any open set is the union of the inverse images of basis elements contained in the set, each of them as a product of two open sets has the inverse image equal to the product of the inverse images of the two open sets, and since and are continuous, the inverse images of the two open sets are open and their product is open.

11 , is open in implies and there is a basis neighborhood in containing such that it lies in . It follows that , and, therefore, is open.

12 (a) It is symmetric, so we check for only. If then and continuous, otherwise it is given by the expression and is continuous as well. (b) if and for . (c) is not closed.

13 Let and be two continuous functions on such that they agree on . Let . In particular, . According to Theorem 18.4 is continuous, and , where . According to Exercise 17.13 is closed in iff is Hausdorff. Since it is closed, is closed. Therefore, . The requirement “if it may be extended” is needed, of course, because not every continuous function can be extended onto the closure of the domain: for example, defined on cannot be extended onto .

Munkres, Section 19 The Product Topology

1 Each of the product is in the corresponding topology, and every basis set (as it was defined in the text) of each topology is a union of basis sets as defined in Theorem 19.2.

2 so that for a given set every element in the set has a basis neighborhood in the topology of the product iff every element in the set has a basis neightborhood in the topology of the product as a subspace.

3 If two points in the product are different, then they have at least one different coordinate, we can separate their values in that space by two open sets and take the product of all other complete spaces and these neighborhoods — the result will be two disjoint subsets of the product open under either topology.

4 The homeomorphism is . Let equals the product of the first spaces. If a set is open in then every its element has a basis neighborhood in which can be represented as the product of open sets, and the product of the first of these sets is open in . Similarly, vice versa.

5 The first part of the theorem relies on the fact that the inverse image of the projection is open, which is true in both topologies. Therefore, the proof still holds, and if a function on the product space is continuous in either topology, then each component of the function is continuous.

6 Suppose that the sequence converges in the product topology. Let be an open neighborhood of in . Then the product of and all other spaces gives an open neighborhood of in the product and all members of the sequence starting from some lies in the neighborhood. This works for the box topology as well. The other direction. Suppose, that the projections of the sequence converge. Take any neighborhood of , it contains a basis set containing . is the product of open sets in each coordinate space, and if we are in the product topology, for all ’s except for a finite number: . For every there is a number such that starting from this number all projections of elements of the sequence into are contained in . If , we choose . For the product topology, there is only finite number of ’s that are greater than , and we can take the maximum of all ’s. All elements of the sequence starting from the maximum of ’s are contained in . For the box topology this direction may fail: there might be no greatest . Some examples are in Exercise 4(b) of §20 (note that in all examples every projection converges to 0).

7 First, we show that under the product topology every point is a limit point of some sequence of points in . Using Exercise 6, in the product topology a sequence converges to a point iff all coordinate projections converge to the corresponding projection of the point. For in define a sequence of points such that the first elements of are equivalent to the first elements of and all others are zero. Then, clearly the projections of onto any coordinate converge to , and, therefore, converges to , i.e. is a limit point of in the product topology, and (see also Example 7 of §23). Second, we show that under the box topology every point outside of has a disjoint open neighborhood. In the box topology, if a point then there is a sequence such that for all , and for each there is a neighborhood of that does not contain . For all other coordinates (not in ) let us take . The product of all is open in the box topology, contains but does not intersect . Therefore, in the box topology is open, and .

8 Obviously, is bijective (as a product of bijective functions). Also note, that the reverse function has the same form with different coefficients. So, we need only to prove that is continuous. Any point in an open set in the product has open basis neighborhood in the set, which is a product of open sets in . The preimage of each such set is open in as well, and the preimage of is . Therefore, any point in has an open neighborhood in this set, and therefore, the set is open, is continuous. This works for both product and box topologies.

9 Given the definition of the product on page 113, the product is not empty if the choice axiom holds. Vice versa, if for any collection of sets the product is not empty, then any element in the product is the choice function.

10 (a) There exists at least one topology such that every function is continuous: namely, the discrete topology. The intersection of all such topologies is the coarsest topology such that every function is continuous in it. Another way to see this is as follows. Suppose there are two such topologies that are not comparable. Then in each topology there exists an open set which is not open in the other topology. Let us take an intersection of these two topologies. It is coarser then both topologies, and the inverse image of any open set in is open, as it must be open in both topologies.

(b) Every set in must be open in . So, every topology such that every given function is continuous in it must contain this collection of sets, and their arbitrary unions of finite intersections. The coarsest such topology is generated by as a subbasis.

(c) Let . Then if is continuous, every is continuous. If each is continuous, then for every open in , is the union of the finite intersection of some subbasis elements in , and for each such subbasis set there is a function and an open set in such that this subbasis set equals to . Therefore, is the union of the finite intersections of and, therefore, it is open in .

(d) Let and . Take such that . Since is open there is a basis element in such that and where is open in (all ’s are without loss of generality assumed to be distinct). For all other ’s assume . Then, and . This set is open in the subspace topology (as only finite number of ’s are distinct from their spaces) and . Therefore, is open in the subspace topology.

Munkres, Section 20 The Metric Topology

1 (a) The basis elements are squares turned by 45 degrees (right angle “rhombuses”). It is a distance by inequality on page 122. The equivalence can be shown the same way as for the “squares” and “balls”. Every such rhombus is contained in the square with the same size. Every such rhombus of size contains a square with size . In other words, it follows from .

(b) It follows from .

2 Let if and otherwise. Any ball under this metric is either a vertical interval open in the dictionary order topology or the whole space. For any open set in the dictionary order topology and any its point there is an open vertical interval centered at the point and contained in the set. There is an open ball in this interval that is centered at the point.

3 (a) and . Let is in or . Let be . Take any point within basic open set . Then and is contained in the same set as .

(b) If is continuous, then for any fixed : is continuous (Exercise 11 of §18). Therefore, every basic -open set is open in .

4 (a) The finer is the topology of the range, the “harder” it is for a function to be continuous. So, for each function we may specify the coarsest topology out of all three given such that it is not continuous in it. For it is the uniform topology, so that is continuous in the product topology only, and for and it is the box topology (they are continuous in both other topologies). is open in the box topology, but all three functions give the inverse image equal to . For the uniform topology the ball has an open inverse image for and only. At the same time, as it is easy to see, all three are continuous in the product topology (for any basis element only finite number of open sets will define some range for the size of the ball centered at any point in the inverse image, or, alternatively, apply Theorem 19.6).

(b) The finer is the topology, the “harder” it is for a sequence to converge. If converges in a metric space, it converges to (any other point will give a non-zero distance to the members of the sequence starting from some index). In the product topology (metricized) it converges as , and, therefore, any ball centered at 0 has all elements of the sequence starting from some index. But in the uniform topology the distance always equals 1. So, converges in the product topology only. converges to in the uniform topology, but does not converge in the box topology, as, for example, there are no points of the sequence in (actually, this show that it does not converge to or any point in this open set, but it obviously does not converge to any other point as well). converges to in the uniform topology, but not in the box topology (the same set shows this). converges to in all three topologies, as it clearly converges in the finest one — the box topology.

5 The finer is the topology, the more closed sets it has, therefore, the closure under the finer topology is a subset of the closure under the coarser topology. Exercise 7 of §19 shows that the closure of under the box topology is the set itself, and under the product topology is the whole space. So here the answer can be either one of those or anything in between. Let be the set of all sequences of real numbers that converge to . Note, that . If a sequence of real numbers does not converge to 0 then there is a positive and an infinite subsequence such that each point in the subsequence is at least from . This implies that the uniform ball around of the size, for example, does not contain any point in . Therefore, is closed and all points outside are not in the closure of . On the other hand, for any and : has a point in . So, the closure is the set of all sequences of real numbers converging to zero.

6 (a) is in but not in the ball. (b) A point in (a) has no ball centered around it that is in . (c) The distance between any point in and is less than . So the RHS is in the LHS. Any point in the ball is such that the distance to is , and, therefore, it is in .

7 , so if we find the conditions on ’s and ’s such that maps open sets to open sets we can apply this conditions to ’s and ’s to find the conditions under which is continuous. The combination of these two sets of conditions will give us the conditions on such that it is homeomorphism. Let be open, then it is a union of balls. . So, it is a homeomorphism. One could try to apply the following logic to prove that it is a continuous function. Since we have already shown (in the exercise 8 of §19) that it is a continuous mapping in both coarser product and finer box topologies, it must be continuous in the uniform topology. The problem here is that we change the topology for both the domain and the range. In this case if a function is continuous when both its domain and range are in a coarser topology and it is continuous when they both are in a finer topology we cannot conclude that it is continuous in the given topology. For example, and is a continuous function when both the domain and the range are in the finite complement topology and, of course, in the discrete topology, but it not continuous in the standard topology.

8 (a) Let . Then it is open in the box topology, and .

(b) The finer is the topology in the space, the finer it is in the subspace. So, given (a) and the fact that uniform topology is finer then the product topology, to show they are different in the subspace we need only to find an open set in the box topology of the subspace that is not open in the -topology, and so on. The set we used in 4(b) which is an infinite product of intervals is open in the box topology, and its intersection with is open in the subspace. , but for all , has points not in the intersection. Now, is open in -topology, but for any there is a point such that it is in and within the ball of size centered at 0 in the uniform topology, but not in . Finally, for any ball , , there is a sequence such that it lies within and -ball centered at 0 in the product topology, but not in (if was chosen sufficiently large).

(c) (Hilbert cube) The box topology is strictly finer than the -topology: is open in it, and in any there is a sequence that is not in the set. The all three other topologies are equivalent on the cube. To show this we show that the product topology is finer than the topology. Intuitively, this happens because every unbounded set in the basis element of the product topology is now bounded by a sequence such that it converges in squares. For any consider the ball . Let us take an open set in the product topology of the form . Then the square of the distance between and any point in the set in the topology is no greater than which can easily be made less than for sufficiently large and sufficiently small 's.

9 (a) Obviously. (b) . For the opposite value of we get . Therefore, . Now take the values in the hint, and it is done. (c) and both are positive. (d) and is iff . The triangle inequality holds.

10 (a) Partial sums are bounded by the product of norms. (b) is obvious, and follows from 9(c) (partial sums are bounded). (c) It is. The sum converges by (b), nonnegative, and zero iff , the triangle inequality holds by 9(c) (using for partial sums).

11 On is well-defined, nonnegative, zero at only, symmetric, strictly increasing and bounded. To show the triangle inequality we use the hint. First, is strictly decreasing in as is strictly concave. Therefore, . Using the fact that and is increasing: . So, is metric. Now equivalence. Since and are continuous, is continuous in -topology as a composition of two continuous functions (using Exercise 3(a)), and vice versa, is continuous in -topology, which means the topologies are the same (Exercise 3(b)).

Munkres, Section 21 The Metric Topology (continued)

1 Let be a ball centered at given by the metric restricted on . with the restricted metric has topology generated by for all . as a subspace has a basis element for all . For every : . So the topology of generated by the restricted metric is coarser then the topology of as a subspace of . Consider any non-empty set for . Let and . Then . Therefore, is open under the restricted metric.

2 By the properties of metric it is injective (therefore, there exists bijective from onto ). The inverse image of any open ball in as a subspace of (which, by Exercise 1, has the restricted metric of ) is an open ball in . Similar for .

3 (a) It is well-defined as the product is finite. All properties including the triangle inequality are obviously satisfied (similar to the proof on page 122).

(b) is a metric on (Theorem 20.1), and, therefore, is a metric. It is well-defined and all the metric properties hold (similar to Theorem 20.5).

4 For : . For in the ordered square : if then , if then , if then , if then , and, finally, if then .

5 Exercise 6 of §19 shows that and we may now apply Theorem 21.3 and Lemma 21.4.

6 For every : , but for any and points sufficiently close to : (something like , which can be proved by induction, might help).

7 Once the notion of the uniform metric on is clarified, the rest is an easy implication. By definition,
(we need the minimum to bound the metric so that it is well-defined).
Now, it is clear that if converges uniformly to then starting from some : . And vice versa.

8 is continuous, by Theorem 21.6, therefore, there is an open neighborhood of such that . Let be such that for : and for all . Then, by the triangle inequality, for .

9 (a) It’s unimodal, the mode at , , and for : as well. (b) The value .

10 The functions are continuous by Lemma 21.4 and as the composition of continuous functions. Then we may apply Theorem 18.1.

11 (a) It converges to the supremum of the set , as can be easily shown as the sequence is non-decreasing.

(b) For sufficiently large both partial sums are sufficiently close to their limits so that the linear combination of the partial sums sufficiently close to the linear combination of the limits.

(c) Both sequences of partial sums in the hint are non-decreasing and bound, so that, by (a), they converge, and by (b), their linear combination also converges.

(d) By (c), converges point-wise to some function . Now, for : . Therefore, uniformly for all (if that was not true than for some sufficiently large the previous inequality would not hold as well). Now, note that for any we can find such that for : .

12 (a) If a sequence converges to then -coordinate converges to and -coordinate converges to and starting from some point they are within a ball of size of their limits. Using the hint, the sum is within a ball of size .

(b) Similar to (a) but taking another from the hint.

(c) The inverse image of is (we may consider finite intervals only): if they have the same sign, if they have different signs, if , and if .

(d) The operation of taking the opposite element is continuous. So, the composition of the sum with it is continuous. Also, the composition of the multiplication with the operation of taking reciprocals is continuous on .

Munkres, Section 22* The Quotient Topology

1 , and their union are open, and other proper subsets of are closed.

2 (a) We can reformulate as follows: if a continuous function has a continuous right inverse then it is a quotient map. has a right inverse, therefore, by exercise 5(a) of §2, it is surjective. Since it is also continuous, all we need to show is that it maps open saturated sets to open sets. Let be open. Now, is open as is continuous.

(b) The inclusion map is a continuous right inverse for the retraction which is continuous by definition.

3 Instead of consider the following function: let be the retraction .Then it is a quotient map. There is an obvious homeomorphism between and . Moreover, . Therefore, is a quotient map as well. Now, if were open or closed, so would be . But is not open and is not closed.

4 (a) . Let . It is a retraction, therefore, a quotient map (see 2(b)). There is an obvious homeomorphism between and .

(b) . Similar to (a).

5 open in , open, imply is open in , is open in .

6 (a) Remember, that set is closed in (not in ). Therefore, every one-point set in is closed, and it is a -topological space. At the same time, there is no neighborhood of in that does not contain points in . Therefore, any neighborhood of in intersects any neighrborhood of some points in .

(b) (It is important here that is not an open quotient map.) is not Hausdorff, therefore, the diagonal of is not closed (every neighborhood of contains : §17, Hausdorff Spaces). At the same time, its preimage is the diagonal in union , and it is closed as is Hausdorff (the diagonal is closed) and is closed in .

Supplementary Exercises*: Topological Groups

1 If it is a topological group then is continuous as a composition of continuous functions. If is continuous then it is continuous in both variables (§18), and, in particular, and are continuous.

2 (a) (b) (c) The spaces are Hausdorff, the operations are continuous (§21, exercise 12).

(d) It is Hausdorff (as a subspace). If and then . For a given open the set is open (needs some work).

(e) It is Hausdorff (as a subspace). The operation of multiplication can be represented as a product of compositions of addition and multiplication. This should help.

3 Let . is a subgroup iff , which is true as ( is continnuous and also Theorem 19.5). Now, if (or ) is a subgroup, then the restriction of on is continuous.

4 They are bijective and continuous. Their inverses are continuous as well. For every and : or .

5 (a) iff , therefore, it is a bijection. Now, , therefore, where is the quotient map. Using Theorem 22.2, since is a quotient (a composition of a homeomorphism and a quotient), it induces a homeomorphism . Moreover, .

(b) is closed, is a homeomorphism, therefore, is closed in .

(c) is open, is a homeomorphism, therefore, is open in .

(d) Using (b), is a -space. Now, let . Using exercise 1, we need only to show that is continuous. Similar to (a), using the fact that the subgroup is normal, (indeed, for this to hold we need ), therefore, . and are continuous, moreover, is an open quotient so that is an open quotient as well, and using Theorem 22.2, induces a continuous function .

6 with addition by modulus. Or, just .

7 (a) and are continuous, therefore, there is a neighborhood of such that and such that . Therefore, .

(b) We need a symmetric neighborhood of such that for no : or . In other words, we need a symmetric neighborhood of such that does not contain . is a -space, therefore, is closed, and there is a neighborhood of such that it does not contain . Using (a), we construct .

(c) We need a symmetric neighborhood of such that for no and : or . In other words, does not intersect , which is closed ( is a homeomorphism). Let be a neighborhood of such that it does not intersect and be a symmetric neighborhood of such that (see (a)).

(d) is closed, therefore, is a -space (5(b)) and is closed. A closed subset of is an image of a closed saturated subset of . Let be a saturated closed subset of that does not intersect . Using (c), let be a symmetric neighborhood of such that does not intersect . Then, since is saturated, does not intersect . Suppose, . Then for some elements , which contradicts . Using 5(c), is an open neighborhood of disjoint from , an open neighborhood of .


Permanent link to this article: http://dbfin.com/2011/02/2000-munkres-topology-solutions-chapter-2/


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  1. bara

    nice work

    1. Vadim


  2. tesfalem

    the solution for (a,b) is homeomorphic to (0,1) is not clear. so please help me mean the verification.

    1. Vadim

      That will be easier for me to find the problem you are asking about if you provide the section and problem number. Thank you.

  3. Molly

    Would it be possible to make those tabs on the right side a bit smaller or move them to the left side? They cover a small portion of your lovely proofs.


    1. Vadim

      Indeed, the site looked wierd on smaller horizontal resolutions. Just fixed the issue, sorry for the delay. Please let me know how this works for you now.

  4. AnthonnyAG

    Hi! You have been really helpful. You give the answer but let the details to the reader, that’s a good thing. Thanks.

  5. Conan

    Thank Euler your website is back. I’m going through Munkres and your answers are very helpful!

  6. Neeraj Bhauryal

    please check solution to problem 9 in section 17

  7. Neeraj Bhauryal

    sorry i mean problem 10 in section 17

    1. Vadim

      Looks good to me. Why?

      1. Neeraj Bhauryal

        Yeah its correct

  8. farshad

    in exercise 7 in section 19: in first part where you prove forproduct topology could you explain more about the sequence Yn and that part that you write ” the first k members are equivalent to xand allothers are zero”?
    and in secont part of it could you please explain about “that does not contain 0 – the product of all such intrvals and all other spaces” what you mean OTHER SPACES

    1. Vadim

      Hey, I have elaborated in more detail in the solution. Hope it helps.

  9. zahir

    it is a very good job and it help me to solve problems in my work@ thank you very much

  10. rojer

    Section 18, 2 is incorrect. Any point must be a limit point or an isolated point. For the constant function f(x)=c, c being an isolated point requires the preimage of {c} to be open, as f is continuous, so {x} must be isolated and therefore not a limit point.

    1. Vadim

      The preimage of is the whole space if is constant.

      The question is whether the image of a limit point is necessarily a limit point. My answer is No, and to support it one needs to provide an example of a continuous function such that the image of a limit point of is not a limit point in . I suggest looking at an example where is constant. To elaborate more on this, consider a function , this function is obviously continuous, and every point in is limit, but their image is not.

  11. hengxin

    Good job! Thx.

    However, I don’t understand your answer to exercise 17 in section 17. For example, how could the limit points of be ?

    1. Vadim

      I think you are referring to exercise 18 in section 17. This is a bit tricky exercise but becomes quite easy once you can imagine the picture and figure out couple things about it. And remember that we are working in a dictionary order topology.

      Let’s first look at a point where < < . For each such point < < for some < < < < , so there is an (arbitrary small) open interval on the same vertical line containing the point. Hence, it is a limit point of a set only if it is a limit point of the set restricted to the interval (the set has points different from and arbitrary close to it).

      Now let's look at a point where < . The point is the lowest point on its vertical line, so all smaller points are to the left of it. For any point : < iff < . Now, the most important part for understanding: any set not containing point has it as a limit point iff either it contains points on the vertical line arbitrary close to or for every < it has some point on a line where < < . This is because in order to be a limit point of , every open interval < < has to contain a point of , and if there are no points of between and , there has to be a point between and , i.e. points in being to the left but arbitrary close to the vertical line (note, that in this case -coordinate of such points does not matter).

      Similarly for points where < . It is a limit point of a set iff either it contains points , < , arbitrary close to or for every > there is a point such that < < .

      For example, sets and both have all points described in the last two paragraphs as their limit points.

      The last two points to consider are the smallest point and the largest one . These are limit points of iff has different points arbitrary close to them on corresponding vertical edges.

      1. hengxin

        Thanks for your efforts. Following this exercise and your detailed comments, I know better about both the limit points and the order topology now.

  12. hengxin

    Nice proof for the exercise 13 in section 18!

    1. Vadim

      Thanks. I looked at it again, and, indeed, it looks nice. If I were to solve it now, I might have chosen a more straightforward way to do it.

      I also posted a reply to your previous comment.

  13. hengxin

    For exercise 11 in section 21 “The Metric Topology”:
    It is quite nice to prove that the bounded metric $d’$ gives the topology of $X$ by making use of the conclusion of exercise 3 (b). However, I don’t understand 3(b) well. Therefore, I am confused about this proof.

    In order to show that the two topological spaces $X$ (induced by $d$) and $X’$ (induced by $d’$) are the same. We can show that (1) $X’$ is finer than $X$ and (2) $X$ is finer than $X’$. To prove the former one (the latter one goes similarly), using exercise 3(b), we shall first justify its premise (i.e., the “if ” part) $d : X’ \times X’ \to \mathbb{R}$ is continuous. I don’t know how this is achieved in your solution. Would you mind offering me some details?

    1. Vadim

      I believe it is Exercise 11 in Section 20.

      In 3(b) you have a different topology such that is continuous. To show that in this case is finer, you need to show that every ball is open in , and this will be enough because those balls form basis of the -topology on . Now, a ball around a fixed point is defined as points with the distance from less than , i.e. the preimage of an open in interval under the continuous function .

      In Exercise 11 we show that two topologies are equal using Exercise 3(b). and are both continuous (the latter exists because is strictly increasing), moreover, and are both continuous (the first in -topology, and the second in -topology), therefore, the metrics give the same topology on .

    2. Vadim

      I have also changed wording in Exercise 3 and 11, so hopefully it is more clear now.

      1. hengxin

        Thanks for that.

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