«

»

Print this Post

2000 Munkres
Topology: Solutions > Chapter 4 Countability and Separation Axioms

Important!

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 30 The Countability Axioms

1 (a) Let be a countable basis at . Then for each there is a neighborhood of such that does not contain . Therefore, there is that does not contain . Thus, .

(b) Given any and there must be a neighborhood of that does not contain , therefore, the space is a -space and every one-point set is closed. We can start with some first-countable -space. By (a), every one-point set in such a space will be a set. Then, if we take a finer topology, it will still have the property, however, we may loose the first-countability if the space is large enough. For example, if we take then it is a metric space in the product topology. Therefore, every one-point set is the intersection of the countable collection of balls of radius centered at the point. Now, if we switch to the box topology, then it seems not to satisfy the first-countability axiom (take a countable collection of neighborhoods of a point and construct a neighborhood by taking the product of open proper subsets of ).

Also, another example would be any countable -space which is not first-countable. Such a space will also be separable and Lindelöf but not second-countable. One example of such a space I found by solving Exercise 17.

2 Let us choose as in the hint (whenever it is possible). It is a countable subcollection of which is a basis: for any open and there is an open set containing , an open set containing and an open set containing , thus, does exist.

3 Suppose the number of limit points in is countable. For each point in that is not a limit point take a basis neighborhood that contains this point in only. All such neighborhoods must be different. Contradiction.

4 As in the hint: take a covering by and its finite subcovering. The union over all is a countable basis. Another way to show this is as follows: compact metric implies Lindelöf metric implies second-countable.

5 (a) For each point in the dense take a countable collection of balls of radius centered at the point. Let , is open. There is a ball . Find a dense point in where . Then .

(b) Similar to exercise 4 (the first part) with the only difference that the subcovering is going to be countable.

6 is separable but not second-countable (proved in the text). The ordered square is compact (it is a linear continuum and a closed interval) but not separable (take the uncountable collection of vertical open intervals, each must have at least one dense point). Note that both are examples of first-countable but not second-countable spaces.

7 Both are not separable: any countable subset of has an upper bound. Therefore, both are not second-countable. is first-countable: for any such that is not open ( has no predecessor) the countable collection of intervals for all is a basis at ; but not Lindelöf: has no countable subcover. is not first-countable: the added point has no countable local basis (for the same reason: the supremum of countable numbers of infima of open sets will be less than ); but is Lindelöf as it is even compact. This illustrates that a subspace of a Lindelöf space may be not Lindelöf.

8 It is a metric space, therefore, it is first-countable and all other three properties are equivalent. In fact, they do not hold. We show that it is not second-countable. According to exercise 3, if it was second-countable, then every uncountable set would have a limit point. But if we take a set of all sequences of 0′s and 1′s, then it is uncountable but the distance between any two elements is 1.

9 This is pretty much the same as for compact spaces: take any open covering, extend open sets to open sets in , add , find a countable subcovering. Now, the second part asks to show by an example that a closed subspace of a separable space need not be separable. Indeed, is separable but its “inverse diagonal” is uncountable and discrete in the subspace topology.

10 Fix any . Consider all points that equals at all but finitely many coordinates and equal some dense points at all others. There is a countable set of finite subsets of coordinates and for each a countable set of points we just defined.

11 If is a continuous map then the dense maps to the dense (each open set contains the image of a dense point in the preimage of the set) and for any covering of the image a subcovering may be obtained as the image of a subcovering of the preimage of the collection. Therefore, if the space is separable, so is the image, and if the space is Lindelöf, so is the image.

12 We just need to show that the image of a basis is a basis of the image of the space. If is a neighborhood of then is a neighborhood of , and there is a basis open set containing , its image is a basis neighborhood of (as is open).

13 Each set in the collection must have a dense point and they must be different as the sets are disjoint.

14 Consider any covering of the product by basis open sets . For every find a finite subcovering that covers (the projection is an open map). Let . Find a countable subcovering of . Then is a countable subcovering of . Indeed, for any point there is such that and also such that . Since , we have .

15 It is the space of all continuous on functions with the uniform metric. We need to show that the space is separable (and, therefore, being metric, it is also second-countable). Consider any finite subset of that includes both end points. There are countably many such sets. Now consider all functions that takes rational values at the points of such a finite set and are linear between the points. They are all continuous functions (use the Pasting Lemma from §18 or the fact that the function is bounded, therefore, the image lies within a compact Hausdorff space and the result from §26 tells us that it is continuous iff its graph is closed) well-defined uniquely by the condition above (all points in a finite set are isolated and for each point not in the set there is the closest point below and another above), and there are countably many of such functions. All we need to show that any continuous function can be approximated by functions from this collection. This is quite easy to show using the results from §27. A continuous function from a compact metric space to a metric space is uniformly continuous. For a given find such that if then . Take a finite set of points of such that the distance between a pair of successive point is less than . For each point find a rational number and construct the dense function . Note that . Then for any we have .

16 (a) The idea (similar to exercise 15): any open set is the product such that only finitely many open sets are not equal spaces. We can always separate the indexes of these spaces by some rational points. So let’s take all finite subsets of rationals in containing 0 and 1 (countably many), each such a subset generates a finite number of half-open intervals of the form [), and then take products of rational points such that they are the same across each interval (countably many).

(b) Define as in the hint for a fixed interval. Note that all are different (indeed, for : must have a dense point but lies within only). Therefore, is an injection and .

17 It is not first countable for the reason exactly the same as in Exercise 1(b). Therefore, it is not second-countable. It is countable, therefore, separable and Lindelöf.

18 (Occasionally we use some results from the Supplementary Exercises of Chapter 2.) So, suppose is a countable basis at . If has a countable dense subset , then we show that form a basis. Indeed, for any : is a neighborhood of , there is a symmetric basis neighborhood of such that , there is a dense point in the neighborhood of , and . Now, suppose that for every covering of there is a countable subcovering. For each take the covering of the space by and find a countable subcovering. The union over all forms a countable basis. Indeed, take . Choose a symmetric basis neighborhood of as before, find such that is a basis neighborhood of , then .

Munkres, Section 31 The Separation Axioms

1 For and take first a neighborhood of such that and then a neighborhood of such that .

2 Similar to 1.

3 Let where or can be infinite. Let be a point in if there is such a point, otherwise (note that in this case , therefore, no points below are limit points of ). Similarly choose in or if the interval is empty. Then and .

4 If a space is Hausdorff then it is Hausdorff in a finer topology. There is no such relation for the two other properties. Indeed, is normal and regular, while is not even regular. At the same time every space in the discrete topology is normal.

5 The set is the preimage of the diagonal of (which is closed as is Hausdorff) under the continuous (Theorem 19.6) function .

6 Take a closed set and an open set . is a closed set in an open set. Find an open neighborhood of such that . is closed and . Indeed, implies for some ( is surjective) but for no implies . Also, implies ( is surjective). So, where is open. Moreover, is closed and contains . Therefore, .

7 (a) In Exercise 12 of Section 26 we established that if is closed and containing is open, then there is some neighborhood of such that . More generally, we can state the same fact for any subset instead of just one point (take the union of all such neighborhoods for each point in ). Take two different points , their preimages are disjoint compact subspaces of a Hausdorff space and can be separated by open neighborhoods . Now find neighborhoods of and such that their preimages are in and . Since and are disjoint, so are the neighborhoods of and (we, probably, need the fact that is surjective here). (How did we use the continuity of ?)

(b) and closed are in . is compact and is closed. Since is regular, for each point in we can separate it and by their neighborhoods. Since is compact we can take a finite number of neighborhoods of its points such that their union still covers , the corresponding finite intersection of neighborhoods of is an open neighrborhood of disjoint from . Now, we use the same fact from Section 26 to find neighborhoods of and such that their preimages are in and , correspondingly. Since is surjective and and are disjoint, the neighborhoods of and are disjoint as well.

(c) Let . is a non-empty compact set. Each point of it has a neighborhood that lies within a compact subspace of . All such neighborhoods cover and there is a finite subcovering , a neighborhood of . The corresponding finite union of compact subspaces of is compact and contains . Once again, we use Section 26 result to find a neighborhood of such that its preimage is in . Therefore, the neighborhood lies within ( is surjective), which is compact as the continuous image of a compact set.

(d) Let be a countable basis of . For each , a finite subset of , define and be the union of all preimages of open sets in that lie within : where is the largest open set in such that its preimage is in . is not empty: take any point in and find its neighborhood such that its preimage is in (we use the same fact from Section 26). There are countably many sets in the collection of all sets . We show that they form basis for the topology of . Take , is open. Let (compact), (open). For each point in choose a basis neighborhood contained in , is compact, choose a finite subcovering of contained in . Note that there is a neighborhood of such that its preimage is contained in , therefore, and is a basis neighborhood of contained in . Done.

8 Let be the quotient map. Then , compact as the continuous image of the compact space . Suppose now, that is closed. We show that is closed. This would imply that is closed, therefore, is a closed map, and we can use the previous two exercises to prove what we asked to. Let . For each , , is closed, therefore, there is some neighborhood of that does not intersect . Since is continuous, we can choose some basis neighborhood of such that does not intersect . The union of all covers . Since is compact, take a finite subcovering, the corresponding finite intersection of is a neighborhood of such that does not intersect . Therefore, does not intersect .

9 (a) For every irrational point there is a basis neighborhood contained in , therefore, if then . The rest of , i.e. the rational points, is a countable set of points.

(b) According to the result of the Exercise 5 of §27, since has non-empty interior in being compact and Hausdorff, one of the closed sets in the union must have a non-empty interior in as well. Therefore, there is such that has non-empty interior, and contains a non-empty interval (a,b).

(c) I show a little different result: for found in (b) there is a non-empty interval and such that for every , . Take where and are such that (found in (b)). Let . is not empty. For each and there is a point . Therefore, .

(d) Given (c), for every point and every neighborhood of there is such that . Therefore, if , is an open set containing , then intersects .

Munkres, Section 32 Normal Spaces

1 Take two closed subset in the closed subspace . and are closed in , therefore, they can be separated by disjoint neighborhoods and . Then and are disjoint neighborhoods of and in .

2 Take a point and a point or a closed set contained in . For every fix . For every let be the product of with for all . For a subset let . If the product is Hausdorff, and can be separated by neighborhoods and . Then and are open (the projection is an open map) and disjoint (if there is a common point then belongs to both and ) neighborhoods of and . Now, if the product is regular, then there are disjoint neighborhoods of and , and their projections into are open and disjoint (for the same reason) neighborhoods of and . If the product is normal and is a closed subset of that does not intersect then and can be separated and their projections into separate and .

3 Theorem 29.2: If is Hausdorff then it is locally compact iff for every and its neighborhood there is its neighborhood such that is compact. Lemma 31.1: If is then it is regular iff for every and its neighborhood there is its neighborhood such that . Therefore, if is Hausdorff and locally compact then it is and regular.

4 The only difference from the proof of the Theorem 32.1 is that we construct a countable covering using the Lindelöf property of the space instead of its countable basis.

5 Yes. Yes. Both are metrizable.

6 Suppose is a pair of separated subsets of . Then is an open subset of that contains both and . . Thus, and can be separated by open neighborhoods in . Since is open, these neighborhoods are also open in .

Take a set and two disjoint subsets closed in . . Similarly, . Therefore, and can be separated by neighborhoods in and their intersections with separate and in .

7 (a) Yes. A subspace of a subspace of a completely normal space is a subspace of the space, and, therefore, is normal.

(c) Yes. If a set is well-ordered then every element has a successor and is a basis at where or . Therefore, as in the proof of the Theorem 32.4, for a pair of separated sets we can cover each set with such neighborhoods that do not intersect the other set. Moreover, the neighborhoods belonging to one set do not intersect the neighborhoods belonging to the other set.

( For any ordered set such that either each point has a successor or each point has a predecessor the proof above works as well: for the latter case we use basis neighborhoods of the form . Also, the next exercise states that every linear continuum is normal: a linear continuum is an example of an ordered set such that it satisfies the least upper bound property the same as a well-ordered set does but every point "in the middle" has no successor or predecessor. Note that even though we know that an arbitrary ordered set is normal, its subspace topology is in general finer than the order topology on the subspace, therefore, we cannot immediately conclude that every ordered set is completely normal. However, it is true as well. )

(g) Yes. Similarly to (c), we can prove that is completely normal. Indeed, the proof of the Theorem 32.4 is extremely similar to the one of Example 2 of the previous section: both use the fact that there is a basis at with sets of the form or and both use only the fact that every point in one closed set has such a neighborhood that does not intersect the other set. Then coverings by such basis sets are disjoint automatically. So, as in (c), we obtain the disjoint open neighborhoods for any pair of sets such that neither one contains limit points of the other one.

(b) No. is completely normal (see (g) above), but is not even normal.

(d) Yes. Every subspace is metrizable as well, therefore, normal (Theorem 32.2).

(e) No. See Example 2.

(f) Yes. Every regular second-countable space is normal (Theorem 32.1). Also every its subspace is also regular and second-countable (see Section 31). Therefore, every regular second-countable space is completely normal.

8 (a) is non-empty and closed, is a component of . If then is empty otherwise it is not. So, suppose it is not. Theorem 24.1 tells us that and all rays and intervals in are connected. Then and are in the same component of iff does not intersect . If is bounded below by a point not in , then let be the greatest lower bound of . Since is closed, (otherwise, let , , i.e. is not the largest element of , and every basis neighborhood of has a from where , but then there is an element , i.e. cannot be an interior point of ). Hence, if there is a component containing a lower bound of then it is of the form . Similarly, if there is a component containing an upper bound of then it is of the form . Now, suppose that is bounded below and above by points in . Let be its greatest lower bound and be its least upper bound. Neither nor belongs to (essentially, for the same reason). But for every there are points such that , therefore, . Hence, .

(b) Suppose is a limit point of . If is in a component then, using (a), the component is a neighborhood of that contains only one point in , therefore, since the space is Hausdorff, must be equal to that point. Suppose now, . Then and each neighborhood of must contain infinite number of points in . Suppose are such points. lies within an interval with endpoints in and that contains neither nor . Therefore, is a limit point of , but is closed, so . This contradicts the assumption that they are disjoint. Similarly, cannot lie within . To sum up, only points in can be limit points of , i.e. is closed.

(c) Using (a) and (b) we conclude that the components of are open intervals or rays. Then is an interval that does not contain points in . Suppose , and . Let be the greatest lower bound of points in . Then, as before, and contains no points in . Let be the lowest upper bound of points in . Then , and has no points in . Therefore, it is a component of with the endpoints in both and , and must have a point in . Contradiction.

Conclusion. For choose those components of that contains points in (they do not contain points in ), and for choose all other components. Since intersects neither nor , we have open disjoint sets separating and .

9 We show that , where is uncountable, is not normal.

(a) is a finite subset of , , . Every is open ( is in the discrete topology). If , is open in , then there is a finite set of indexes such that implies . Therefore, and they form a basis.

(b) Let , i.e. may have many coordinates equal to but all other coordinates different. are closed and disjoint. Indeed, if then and does not intersect . For let . Since is injective on , is countable. Therefore, there is uncountable set of indexes such that for , i.e. cannot be a point in any other .

Suppose and separate and and derive a contradiction.

(c) Essentially, what we do here: we want to construct a sequence of points in such that they “converge” eventually to a point of the form: 1 for all indexes except a countable subset of indexes for which it takes values 1, 2, 3, …. We take the first point equal to 1, then for the first indexes in a countable sequence of indexes we define values from 1 to (), then do the same for the first indexes (), etc. Moreover, we want to define the sequence of indexes and the sequence of “steps” such that . Note that takes values different from 1 on the set while includes the next step as well. Ok, so, for all and . Given and we first define such that (this is possible because and we can define ) and then just define by the expression.

(d) Now, given the sequence of points and the sequence of points , we define a point in equal to 2 for all points not in and for . , so there is a finite set such that . Since is finite, there is such that . Let for , for and for all other indexes. Then for all either and or and . For all either and or and . Therefore, .

10 In the Supplementary Exercises (exercise 7(c)) of Chapter 2 we have shown that is regular. The previous exercises shows that it does not have to be normal.

Munkres, Section 33 The Urysohn Lemma

1 , . So, if , for , , for , . The other direction. If for all then . If for then .

2 (a) Follows from (b), but, in fact, we will prove (b) based on (a). For any completely separable space: take two points, a continuous function that separates them, if some point in the image is missing, can separate: and .

(b) Countable regular implies countable Lindelöf regular implies countable normal. By (a), if it has at least two points, it is disconnected.

3 The distance between a point and a set was defined in §27. It was shown that it is continuous, and iff (exercise 2 of §27). Therefore, for two closed sets the expression is well-defined (the denominator is never 0), continuous and takes 0 on , 1 on .

4 Suppose such exists. must be closed, and, using exercise 1, it is the countable intersection of open sets. The other direction. If is closed and there is a countable collection of open sets such that their intersection is , then let and for every if is defined let . This is possible because the space is normal and is closed. We need to slightly modify the construction in the proof of the Urysohn lemma. Namely, we do not define , we define a sequence of point of starting from 1 such that no is defined before if . So, first we define , then , then some for , then , then some for , etc. This way we can define for all rational points in with the properties required by the proof. Moreover, .

5 Suppose such function exists. For and , and for and , apply the previous exercise. Suppose now we have a pair of disjoint closed sets. Using the previous exercise we construct such that and . Then is what we need. ( Compare to exercise 3: for any closed , satisfies all the properties of a function constructed in exercise 4. )

6 (a) The intersection of -neighborhoods of a closed set (see exercise 2 of §27) is the set itself: since the set is closed, for a point outside it there is a neighborhood that does not intersect the set. Therefore, a metric space being normal is perfectly normal as well.

(b) Let be a pair of separated sets, and be two continuous functions that vanish precisely on and , respectively. Let . It is continuous, and .

(c) We need some completely normal space that is not perfectly normal. This should be a normal space such that there is a closed set which is not set. What types of completely normal spaces we know so far: metric, ordered and regular second-countable. Metric spaces are perfectly normal. In fact, I looked up ahead, and have realised that the Urysohn Theorem states that a regular second-countable space is metrizable. So, this leaves us with the option to find an ordered space with a closed set that is not set. The obvious example is . Its largest element is the one that causes all the problems with the space: it is not separable, it is not Lindelöf, it is not second-countable — all because of the basis at . And, I believe, for the same reason it is not perfectly normal. For each set in any countable collection of neighborhoods of find a basis neighborhood contained in the neighborhood. Then has an upper bound, and the intersection cannot be .

7 is a subspace of a compact Hausdorff space (§29, one-point compactification), which is normal (§32) and, hence, completely regular (§33, corollary from the Urysohn Lemma), therefore, is completely regular (Theorem 33.2).

8 So, what it says is that a completely regular space is “normal” on pairs of closed sets at least one of which is compact. Each point of the compact set can be separated from the closed set by a continuous function (here and ). Then for some fixed , defines an open neighborhood of that does not intersect . covers . We can find a finite subcover , , and take the corresponding finite product of continuous functions . Now, for each : for some , and , therefore, . For every we have . All we need to do now is to transform : find continuous such that for : for and for . This will separate and . Indeed, the finite product of continuous functions is continuous and the composition of continuous functions is continuous. For example, will do.

9 It is sufficient to show that can be separated by a continuous function from a closed set that does not contain it. There is a neighborhood that does not intersect . Consider a linear transformation . We have shown back in the exercises of Section 19 that the linear transformation is a homeomorphism in both the product and box topology on . It seems a more general result is true: a product of continuous functions is continuous (so, that the product of homeomorphisms is a homeomorphism). If , is open in the box topology, then there is a basis neighborhood and iff for all : iff for all : . Therefore, for all which is open in the box topology if all are continuous, . Anyway, it is true for the linear transformation, and if could separate from by a continuous function , then would separate and . Note, that is closed and does not have any points in for if it had some than would have a point in . Now, since is completely separable (as a metric space), there is a continuous function that separates and (remember, that is not open in the uniform metric: exercise 6 of Section 20). Now, is continous in the finer box topology as well, and separates from .

10 In the exercise 10 of the previous section we have mentioned that a topological group must be regular but there is an example of a topological group that is not normal. So, the only question is whether it has to be completely regular or not. Here we show that the answer is positive. Recall (Supplementary exercises, Chapter 2) that there is a self-homeomorphism that maps any element to . Therefore, we only need to show that we can separate from any closed set that does not contain it by a continuous function. Also, for any neighborhood of there is a symmetric neighborhood of such that . So, for every dyadic rational in define by the recursive procedure given in the exercise. Note that the product of any two open sets is open (as the union of open sets of the form a point an open set), and that the product of two neighborhoods of is a neighborhood of .

Let (it is not open but just for convenience to prove the relations below, after that we assume ). Then for all ( for all , and therefore, for any neighborhood of ). Now, , , , , = , , , , . We see immediately that . At the same time, . And (by induction) the latter term is a subset of . Therefore, for every and , (this obviously holds for and outside as well).

Now, given a closed set take a neighborhood of , define as above, and . The only difference from the proof of the Urysohn lemma is in that we define on a dense subset of set instead of the whole set , and for in the dense we proved there is some such that instead of . The first difference does not matter. Now, let . For , since is symmetric, . Therefore, and .

11 (a) The sets cover ((iii) and (iv) is already enough). Now, the non-empty intersections: (i) and (i) is of type (i), (i) and (ii) is (i), (i) and (iii) is (i), (i) and (iv) is (i), (ii) and (ii) is (ii), (ii) and (iii) is (ii), (ii) and (iv) is (ii), (iii) and (iii) is (iii), (iii) and (iv) contains (ii), (iv) and (iv) is (iv) (note, once again, the intersection is assumed to be non-empty).

(b) , and its intersection with is a set containing . Therefore, there are only countably many points (given by the intersection of (ii)-type neighborhoods) in , the set of such that . If does not intersect any of (countably many points) then for every , . Since is continuous and every neighborhood of a point in contains the intersection of some horizontal interval containing the point with , . Now, every open set containing , contains the whole bunch of sets , and the same is true for open sets containing . Therefore, both points a limit points of and .

(c) It is not completely regular because and cannot be separated (and we will show that one-point sets are closed). Now, it is a space: consider points , , two points in and a point in and two points in and a point in . For each point in the list we can easily find a neighborhood that does not contain another point in the list. Now, we show it is regular: every neighborhood of a point contains the closure of another neighborhood.

Let and denote all points to the left (including ) and to the right (including ) of . Then , and are closed. Therefore, any basis neighborhood of (or ) contains the closure of another neighborhood.

is closed (for any point not in it find a neighborhood that does not intersect ). If a point is in then the one-point set is open (the intersection of a small enough horizontal interval and ) and closed. Every basis neighborhood of contains all but finite number of points in , but it is closed as well.

If a point is in then its basis neighborhood is either for some or for some or the intersection of a horizontal open interval with . The first two types of neighborhoods contain a neighborhood of the third type. But the third type neighborhood (small enough so that they do not contain other limit points) is closed as well (the rest of the space is the union of horizontal intersections, some neighborhoods of and , and sets minus finite number of points).

Munkres, Section 34 The Urysohn Metrization Theorem

1 was given before as an example of a space which is Hausdorff but not regular ( is closed but cannot be separated from ). It is also second-countable. Since it is not regular, it is not metrizable.

2 For a metric space being separable and second-countable is equivalent, at the same time if a space is second-countable and regular then it is metrizable, so, an example of a space which is completely normal, first-countable, separable and Lindelöf but not second-countable is necessary and sufficient to answer the question. was shown to be completely normal (in fact, it is perfectly normal) and to satisfy all the countability axioms but one: it is not second-countable.

3 A compact metric space is second-countable (exercise 4 of §30). A second-countable compact Hausdorff space is second-countable and normal (Theorem 32.3), therefore, metrizable (Theorem 34.1).

4 If a space is locally compact Hausdorff then it is completely regular and being second-countable implies being metrizable. At the same time a locally compact metric space does not have to be second-countable (at least I do not remember we proved anything like that). For a metric space being separable, Lindelöf or second-countable is equivalent. So, we need an example of a locally compact metric space which is neither one (it can still be first-countable). I know couple spaces that are first-countable but not separable, Lindelöf or second-countable. Both are Hausdorff and locally compact. The first is , not metrizable (§28, page 181: it is limit point compact but not compact). The second is a discrete uncountable space, metrizable.

5 If is locally compact and Hausdorff then is compact and Hausdorff. is metrizable iff it is second-countable. So, if is metrizable then is second-countable (note that being metrizable is a stronger condition than being metrizable: the discrete uncountable topology is an example, this is why it did not work in the previous exercise). If is second-countable then it is metrizable (see the previous exercise), but we need to check whether is metrizable. For this we need only to check whether it is second-countable. A countable basis for the topology of will do for as well. We only need to find a countable basis at . Take an open neighborhood where is compact in . is compact in iff it is closed in . Therefore, isopen in and must contain some basis neighborhoods . But we do not know whether is compact for any of these basis sets, therefore, we cannot guarantee that is open in for some . At the same time, we may instead build a countable family of compact sets such that every compact set is contained in a set from the family. We use the fact that is compact and the space is Hausdorff and locally compact. Consider the countable family of all basis open sets such that their closures are compact. Since the space is Hausdorff and locally compact, every point has a neighborhood in (Theorem 29.2). Therefore, covers and some its finite subset covers as well. The corresponding finite union of closures is compact (and closed) and contains . Therefore, we may take as basis neighborhoods of the complements of all finite union of compact closures of basis sets of . Something like that.

Summary of 4 and 5: a one-point compactification of a locally compact Hausdorff space is metrizable iff is second-countable.

6 defined as in the proof of the Theorem 34.1 maps to . It is continuous as the range is in the product topology. It is injective because is a -space (for a pair of points there is a function in the family such that it maps the points to different values). We need to show that maps open sets in to open sets in the image. For find an index such that . Then the set of points such that is an open neighborhood of in and its intersection with the image of is open in the image. Moreover, if then and , hence, .

7 Let . Let be an open set containing such that it is metrizable in the subspace topology. Then is locally compact (Corollary 29.3) and Hausdorff. Let be a neighborhood of in such that is compact. Then is a compact Hausdorff metrizable subspace containing . Using Exercise 3, we conclude that the subspace topology of is generated by a countable basis. is open in and second-countable as well (in the subspace topology). Now cover with such neighborhoods for all points and find a finite subcovering . Consider the finite union of all countable subspace bases. Suppose . Then there is containing . is a neighborhood of in , and it contains a basis sub-neighborhood that is open in and belongs to . So, is second-countable, therefore, according to Exercise 3, it is metrizable.

8 As in the previous exercise we want to cover with a countable collection of open subspaces such that each one has a countable basis. For take open metrizable containing , find open such that . is a closed subspace of a Lindelöf space, i.e. Lindelöf. It is also metrizable, therefore, being Lindelöf is equivalent to being second-countable. Hence, is an open second-countable (in the subspace topology) neighborhood of . Now cover with such neighborhoods, find a countable subcovering, and prove that the countable union of countable bases of all open sets in the subcovering is a countable basis of . Therefore, is regular and second-countable. Hence, metrizable. We used the regularity twice: to find a neighborhood with the closure within a given neighborhood, and to argue that the space being regular and second-countable is metrizable. If the space is Hausdorff, Lindelöf and locally metrizable then it is not necessarily metrizable. To find a counterexample we need a space which is Hausdorff but not regular. is such a space. It is also Lindelöf but not metrizable. The only question is whether it is locally metrizable. The subspace topology on and is the same as the standard sub-topology, therefore, metrizable. The only question is now whether 0 has a metrizable neighborhood. But is its neighborhood with the same topology as the standard one, therefore, it is metrizable.

9 Each subspace is compact, Hausdorff and metrizable. Therefore, second-countable (Exercise 3). According to the same exercise, we only need to show that is second-countable. If and were disjoint, they would be open as well, and we could take the union of their bases: if then every its neighborhood in has a sub-neighborhood , open in and , and, therefore, it would contain a basis neighborhood open in and . If they are not disjoint, then we do not even know that a basis set in is open in . Let be a basis in . Suppose is a neighborhood of . If then is open in and there is a basis neighborhood containing . Moreover, is open in which is open in , therefore, is open in as well. This suggests that, first, we take all intersections (they are all open in ). Note that if and were disjoint this would be exactly the union of their bases. In either case, this already provides bases for all points not in . Now, suppose . has a basis sub-neighborhood of in . where is open in . . So, for every pair of intersecting basis sets in and choose some open sets in that intersect with the subspaces at these basis sets and take their intersection. We have a countable basis for points in .

Munkres, Section 35* The Tietze Extension Theorem

1 Take the continuous function on the union of two disjoint closed sets equal to 1 for one set and 0 for the other set (it is continuous because both sets are closed and, therefore, open in the union) and extend it continuously on .

2 In this case the approximation by the nth partial sum is and . The partial sums also bounded by .

3 For a metric space (iii) is equivalent to being compact and implies (i) (a compact metric space is bounded) and (ii) (the continuous image of a compact set is compact). (ii) obviously implies (i) (fix a point, the distance from the point is continuous). Now, (ii) also implies (iii): if an infinite subspace has no limit points then it is closed and discrete; a surjective function from it on the set of positive integers is continuous and can be extended to a continuous unbounded function (the space is metrizable, therefore, normal). We need only to show that (i) implies (ii) or (iii). We show (i) implies (ii). The hint was ambiguous for me so I figured out an alternative proof.

Remember from Section 20 that the topology generated by a metric is the coarsest topology such that the metric as a function is continuous. This implies also that if a metric is continuous relative to then is coarser than . We prove the following lemma.

Lemma. Suppose that is metrizable and every metric that generates the topology is bounded. Then every metric continuous relative to is bounded as well.

Proof. Suppose . is bounded. Let . is a metric (easily verifiable) and continuous relative to , therefore, is coarser than . We show that, in fact, . For every , suppose , then , therefore, for every point in there is a neighborhood in contained in the ball, and is finer than . This implies that is bounded, and so is .

Now, take any continuous . And suppose is a metric on . Then is a metric that is continuous relative to the topology on . Therefore, it is bounded, and so is .

4 This is something from algebraic topology, as I understand (and the index suggests that retractions are covered mainly in the second part of the book). But, in fact, there was a fact about retractions (and that was the reason to look up the index) in the section on the quotient maps. A retraction (a continuous map that preserves all points in the image) is a quotient map.

(a) Let , let . There are disjoint neighborhoods and of and . is open in and contains . is open in and contains and . If then and . Therefore, for every , .

(b) The preimage of each point must be open and non-empty, and the union must be the whole space, which is connected.

(c) For take . It is well-defined for all non-zero points, and continuous as a composition of continuous functions. Moreover, it preserves all points on the unit circle. As for the second question, it seems a little more trickier: whether is a retract of . Suppose a continuous function maps continuously onto the unit disc (not necessarily unit circle) and preserves all points on the circle. The restriction of the map onto the unit disc is a continuous map from the disc to the disc that preserves all the points on the circle. It seems that such a function must map some point to the origin (as well as to any other point in the disc). In other words, it must map the disc surjectively onto the disc. How to show this based on what we know? Another way is to show that a continuous map from the unit disc to itself must have a fixed point. Then we take the composition of with the rotation and the new continuous map does not have fixed points on the boundary, therefore, there is no continuous retraction onto the boundary. Yet, another way is just to try to prove that every circle centered at 0 maps to the entire boundary (under the retraction). This way when the circle vanishes to the origin, the origin must maps to all points on the boundary. This seems more promising. Let be the infimum of the set of radii such that every circle centered at 0 with the given radius or greater maps to the entire boundary. If then we show that is in the set and find such that is in the set. To find we, probably, need something like the uniform convergence.

5 (a) can be extended to a continuous , then such that is continuous (in the product topology).

(b) If is continuous, and is a homeomorphism such that is a retract of then there is a retraction and can be extended to a continuous function . Then is continuous.

6 (a) is a homeomorphism, it can be extended to a continuous , then is a retraction.

(b) So we assume is compact and Hausdorff, i.e. normal. If we show that is homeomorphic to a closed subspace of then, since is an absolute retract, is a retract of , and, similar to 5(a)(b), has the universal extension property. The Imbedding theorem provides a way to construct which is an imbedding. For each we define that equals 1 at and vanishes outside of . If is not in the image, then for every there is some such that . Choose a disjoint neighborhoods for and for . The preimages of cover which is compact, there is a finite subcovering. The corresponding finite intersection of is a neighborhood of disjoint from the image. (In other words, the continuous image of a compact is compact and closed in a Hausdorff space.)

7 (a) The quotient space of obtained by identifying points with the same distance to the origin is homeomorphic to and to . This gives us a continuous function from to . The explicit expression for the function will be something like this: where . Every sequence of points converging to the origin maps to a sequence converging to the origin as well, therefore, we can extend to a continuous function .

(b) is homeomorphic to , which is normal. Therefore, using the two previous exercises, it has the universal extension property and is an absolute retract. Being closed in , it is a retract of .

8 One direction: 6(a). The other direction: if a normal space is an absolute retract then it has the universal extension property. The idea, as I understand, is as follows: we consider a continuous function from a closed subspace of to ; we know that whenever is homeomorphic to a closed subspace of a normal space, the subspace is a retract, i.e. there exists a continuous map from the space onto the subspace that preserves all points of the subspace; but we need a continuous function from to that extends ; we consider the space which is the union of the spaces and and group some points into equivalence classes such that every point in belongs to a separate class that also contains the preimage of the point; the union of all classes for all points in is a closed subspace of the quotient space being homeomorphic to ; we show the quotient space is normal and consider a retraction from onto the set of classes of points in ; now we can construct a continuous extension of , as we shall see.

Construct the adjunction space as hinted: the quotient space obtained from by identifying each point with points in (for this we, first, need to specify the topology on : a set is open in iff is open in and is open in ). Let be the quotient map. Let and denote elements of as follows: where and where (if then contains only one element).

First, we show that is normal. Let and be closed in . Let and . are closed in ( itself is closed as it contains all elements from and , both closed in ), therefore, are closed in and we can separate them by a continuous function (Urysohn lemma). Now, is continuous on and . We can extend it onto : . It is still continuous: it is continuous on and (the pasting lemma?). Now we can extend it onto (Tietze theorem). Note that the extended is constant on for all . Therefore, it induces a continuous function on (Section 22) such that it separates and . Hence, is normal.

Now, we show that is homeomorphic to a closed subset of , namely, to the set . Indeed, is continuous and bijective. It maps closed to which is the image of a closed saturated subset of under the quotient map ( is closed in which is closed in ).

Finally, since is a closed subset of the normal space homeomorphic to , and is an absolute retract, there is a retraction such that for every : . Let . is continuous, and maps to . For every : . Therefore, extends .

9 (a) Let be a collection of open subsets of . Then for each , is open in if there are finitely many open sets, and is open in as well. Now we need to show that is a (closed) subspace of . From the definition of the coherent topology on it follows that the topology on is finer than the subspace topology on . We need to show that if is open in then there is some set such that and is open in for every . is a subspace of for every . For let be a set open in such that . Note that for , and for , . Let . Then, for , is open in . For , is open in . Therefore, the subspace topology is finer than the topology on , and, overall, we have that the topologies are equal. To show that is a closed subspace of we can take any point and construct its neighborhood in disjoint from by taking the union of for all .

(b) In the past we had the following fact: if a function is continuous on every space in a collection of subspaces covering the space and either all subspaces in the collection are open or all subspaces are closed and the collection is finite, then the function is continuous on the whole space. Here we have either the infinite union of closed subspaces or the infinite union of open subspaces union one closed subspace . If we knew that where is open in and there is some such that then we could apply the result stated above. However, it might be that every neighborhood of some point in intersects all spaces . For example, suppose and is open in iff for every all also in . Note that every is closed in . Then is but there is no open set in such that it is contained in any . So we show the result directly.

Let be open in . Then for every , which is open in . Therefore, is open in .

(c) Let and be closed disjoint subsets of . Let (it is a closed subspace of ) and be defined on such that and . We assume that . Note that both and are closed and open in and is continuous. For every , given a continuous function such that and , we extend it to a continuous function which satisfies the same property: and . We cannot directly apply the Tietze Extension Theorem because may be not normal. However, is normal. So we can extend to a continuous function as follows: first, we say that and , the function is still continuous ( and are closed in , hence, their intersection with is closed in and the continuity follows by the pasting lemma of Section 18), then, we extend it from the closed subset onto . Now, we define to be equal to on , on and on (it is still continuous by the pasting lemma). For every there is the minimal such that . We define . Note that is not defined for and agrees with for . For we define and for we define . Note, again, that this definition agrees with all . This implies that which is continuous, and by (b), is continuous. Since and , separates and . Hence, is normal.

Munkres, Section 36* Imbeddings of Manifolds

1 Let , open subset of the manifold . There is a neighborhood of homeomorphic to an open euclidean space. Then is open in and is homeomorphic to an open subset of the euclidean space. Let be the homeomorphism. There is a neighborhood of such that . Then is a neighborhood of such that . Now, important! This does not show yet that the space is regular. Consider the following example. Let us take the standard topology on and make a copy of the origin, a new point . The basis for the topology is the collection of open intervals and open intervals containing 0 with substituted for : where . Obviously, this space is a manifold (if we do not require the Hausdorff condition in the definition), however, it is not Hausdorff: we can not separate and (even though for every point and its neighborhood there is a neighborhood of such that , i.e. it is “almost regular”). So, we do need require a manifold be Hausdorff to complete our proof that it is regular.

2 The space is compact and Hausdorff, therefore, normal. Cover it with a finite number of open sets such that each can be imbedded into , let be the corresponding collection of imbeddings. By Theorem 36.1, there is a partition of unity dominated by . Similar to the proof of Theorem 36.2, we construct so that it is continuous and injective.

3 Using the exercise 2, can be imbedded into a second-countable space. Hence, it is second-countable and, therefore, an -manifold.

4 In the proof of the Theorem 36.1 the first step was to show the result of this question for finite families of open set covering . The essential part of the proof is to construct each such that . We need the first inclusion to ensure that it covers , and the last one for later construction of a partition of unity. In the finite case we construct ’s one-by-one such that contains . Note that at the last step covers all of , therefore, all points of are covered. If the covering is countable, and we use the same way to construct sets ’s, then, in general, we cannot guarantee that each point will be contained in some set . However, if every point is contained in a finite number of sets , then for some , does not contain and, since contains , must be contained in . Therefore, the collection of ’s constructed the same way as in the theorem, does cover all of .

[ To construct a counterexample we would need a space such that it is a) normal, b) there is a countable cover by open subsets (not point-finite) such that there is no other cover such that . I have been thinking about the example for a while, but have not figured out one yet. The fact that the space is normal implies, in particular, that the construction above is possible. On the other hand, it seems that the construction is, in some sence, necessary... What I mean is that for any set , a) must cover , therefore, it must cover , and b) it must be such that . So, if there are 's for 's, then they can be constructed using the procedure above. Our only hope is to find a space such that whatever sequence we take using the procedure above, there will be a point such that it is not covered. In particular, when we define such that there must be some points in not covered by such that each is contained in an infinite number of sets . ]

5 (a), (b) It is better to think about the space as described in the solution for the exercise 1: we just add a point and new basis neighborhoods of the point. The intersection of a new basis neighborhood with an old one is either an interval or the union of two intervals or empty. Also, this way it is immediate that the space without the new point is homeomorphic to the real line, and if we substitute the new point for 0, then it is also homeomorphic to the set of real numbers.

(c) It satisfies the axiom: in fact, any pair of points except can be separated by two neighborhoods, while these two points cannot be separated by two disjoint neighborhoods, though each one has a neighborhood not containing the other one.

(d) and are both open, metrizable, and each point belongs to at least one of these open sets. Also, since is second-countable, so is , as where is a basis at the new point.

Supplementary Exercises*: Review of the Basics

Permanent link to this article: http://dbfin.com/2011/03/2000-munkres-topology-solutions-chapter-4/

16 comments

Skip to comment form

  1. shibashis

    thank you very much for publishing this on the web

    1. Vadim

      You’re welcome.

  2. Pu

    I have been looking for these materials for a long time. Thank you very much!

    1. Vadim

      You are welcome.

  3. Abyss

    Thank you for the notes! Very clear. Would you know where I may find help with the supplementary exercises? I’m struggling with #2 and #6 (continuous mapping).
    I have the intuition that:
    - 1 through 4 would be preserved
    - compactness would be preserved
    - Hausdorff might not be preserved
    Thanks in advance!

  4. Abyss

    Hi, I posted a comment a few days ago but it doesn’t appear posted. First of all, thanks for the notes!
    I was wondering if you had given any thoughts on the Supplementary Exercises? I am struggling with #6 and the continuous mapping. I haven’t found much information in Munkres’ notes. Thank you in advance!

    1. Vadim

      Sorry for not responding. I wish I had more time to check it more often. I did solve the supplementary exercises (or most of them), but never had time to actually type it in here. I need to check if I still have the papers. Meanwhile, try to do it yourself. I will post them as soon as I find some time to look into that once again.

  5. Abyss

    Vadim,
    I have tried to solve them. If you give me your email, I’ll be glad to scan/give you my notes if that helps! However I have not been able to prove it for all 17 properties..

  6. Babak

    Dear Vadim

    Your proofs are interesting. But we can exhibit some other proofs that are easier than yours. For instance about 3 of The Tietze Extension Theorem, we can see that d(x,y)+I fi(x)-fi(y) I is a metric on X such that induce the topology on X, also. Therefore by the hypothesis X is bounded under this metric and hence the function ,fi, is bounded. Am I right ?

    1. Vadim

      If I understand you correctly, then what you mean is to show directly that induces the topology equivalent to . This would be the same steps as in the proof of the lemma. I just wanted to state the lemma as a separate result, as the solution looks more structured and useful this way. And it does not seem to make the solution any longer, from my point of view. You still need the idea of using the function , and it is just a matter of personal taste, how you show it is bounded.

  7. anonymous

    Hello, question 8 on section 33:

    How is f^-1 ( { 0 } ) an open set around x if f ( x ) = 0? 0 is not relatively open in [0,1] with the standard topology.

    1. Vadim

      Yes, thanks. It’s got to be something like . I have corrected the solution. Hopefully it looks better now.

  8. farshad

    i have one question:
    you can prove that simply connected is a topological property or not?

    1. Vadim

      Sorry for late reply. I believe “simply connected” is something studied later on in the book in Algebraic Topology, namely Chapter 9. Still, it is a topological property as it is based on the topology only.

      In the case of simply connected spaces, the property requires that the space is path-connected (for any two points there exists a continuous function from the unit interval that connects them) and every self-connected path in can be contracted to a point (for every continuous function from the unit circle into the space there is a continuous extension onto the unit disk). Both requirements deal with the given topologies of the real line and plane, and the topology on the space only (to ensure the existence of required continuous functions).

      A formal proof would require something along these lines: suppose that is simply connected, and consider that is homeomorphic to , i.e. there is continuous function such that there exists and it is continuous as well. In other words, there is a function that serves as a bijection between the open sets of the two spaces. You need to show that in this case is also simply connected. It is a quite straightforward exercise.

  9. mansooreh

    Thanks for good notes. it was useful in other sections either.

    1. Vadim

      You are welcome.

Leave a Reply

Your email address will not be published. Required fields are marked *