Exercises Attempts
Exercise after §30
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(a) A set in a space is a set that equals a countable intersection of open sets of . Show that in a first-countable space, every one-point set is a set.
(b) There is a familiar space in which every one-point set is a set, which nevertheless does not satisfy the first countability axiom. What is it?
(a) A set in a space is a set that equals a countable intersection of open sets of . Show that in a first-countable space, every one-point set is a set.
(b) There is a familiar space in which every one-point set is a set, which nevertheless does not satisfy the first countability axiom. What is it?
(a) Let be a space that is first-countable and and a one-point set containing a fixed . Since is first-countable, there exists a countable collection of open sets containing such that each open set containing contains at least one of the elements of .
Note that by Exercise 17.15, for each pair of distinct points of , there exists a neighbourhood of a point not containing the other and vice versa. This implies that if we take pairs of points of such that one of the points in a pair is , we obtain a family of neighbourhoods of such that does not contain a point . Each of the neighbourhoods also contains at least one of the elements of ; we call it . Gathering such 's together, we obtain a countable family of neighbourhoods of , since it is a subset of . Taking intersection over such 's, we obtain precisely , because for every point , there exists some such that .
Since the choice of is arbitrary, it follows that every one-point set is a set.
(b) The space equipped with the box topology is not first-countable, but every one-point set in it is a set.
To see that a one-point set in is , given a one-point set , where , consider the following countable collection of open sets (under the box topology):
It follows that the intersection of the collection is precisely , due to Archimedean property.
To show that in the box topology is not first-countable, let be a point in and a countable collection of neighbourhoods of . There then exists some basis element that contains , so that for each , and is contained in for each .
Let and , then is a neighbourhood of but for each .
- Show that if has a countable basis , then every basis for contains a countable basis for . [Hint: For every pair of indices for which it is possible, choose such that .]
- Let have a countable basis; let be an uncountable subset of . Show that uncountably many points of are limit points of .
- Show that every compact metrizable space has a countable basis. [Hint: Let be a finite covering of by -balls.]
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(a) Show that every metrizable space with a countable dense subset has a countable basis.
(b) Show that every metrizable Lindelöf space has a countable basis.
(a) Show that every metrizable space with a countable dense subset has a countable basis.
(b) Show that every metrizable Lindelöf space has a countable basis.
- Let be a closed subspace of . Show that if is Lindelöf, then is Lindelöf.
- Let be continuous. Show that if is Lindelöf, or if has a countable dense subset, then satisfies the same condition.
- Let be a continuous open map. Show that if satisfies the first or the second countability axiom, then satisfies the same axiom.
Exercise after §31
- Show that if is regular, every pair of points of have neighbourhoods whose closures are disjoint.
Suppose that is regular, then (by the assumption that is ) is also Hausdorff. Therefore, if we let be a pair of (distinct) points in , there exists a neighbourhood of and of such that .
Applying Theorem 31.1, there then exists a neighbourhood of and of such that and . Since and are disjoint, it follows that and are disjoint as well.
- Show that if is normal, every pair of disjoint closed sets have neighbourhoods whose closures are disjoint.
Let and be a pair of disjoint closed subsets of . By normality, there exists a neighbourhood of and of such that and are disjoint. By Theorem 31.1, there exists neighbourhoods and of and respectively such that and . It then follows that and are disjoint, as desired.
- Show that every order topology is regular.
- Let be continuous; assume that is Hausdorff. Show that is closed in .
8 (reworded). Let be a compact topological group; let be a topological space; let ba an action of on . If is Hausdorff, or regular, or normal, or locally compact, or second-countable, so is the orbit space of , denoted by .
[Hint: See Exercise 13 of §26.]
Exercise after §32
- Show that a closed subspace of a normal space is normal.
Let be a normal space and a closed subspace of . Let and be disjoint closed subsets of . By Exercise 17.2 (in Chapter 2), it follows that and are also closed in . By normality, we obtain two disjoint open sets and in that contain and respectively. It follows that and are two disjoint open sets in that contain and respectively, thus is normal.
- Show that if is Hausdorff, or regular, or normal, then so is . (Assume that each is nonempty.)
- Show that every locally compact Hausdorff space is regular.
Since is locally compact and Hausdorff, it follows by Theorem 29.2 in Chapter 3 that for every point in and neighbourhood of , there is a neighbourhood of such that is compact and . This satisfies Lemma 31.1(a) (although compactness of is not required), so is regular.
- Show that every regular Lindelöf space is normal.
- A space is said to be completely normal if every subspace of is normal. Show that is completely normal if and only if for every pair of separated sets in (that is, sets such that and ), there exist disjoint open sets containing them. [Hint: If is completely normal, consider ].
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Prove the following:
Theorem. Every linear continuum is normal.
(a) Let be a nonempty closed subset of . If is a component of , show that is a set of the form or or , where .
(b) Let and be closed disjoint subsets of . For each component of that is an open interval with one endpoint in and the other in , choose a point of . Show that the set of the points is closed.
(c) Show that if is a component of , then does not intersect both and .
Prove the following:
Theorem. Every linear continuum is normal.
(a) Let be a nonempty closed subset of . If is a component of , show that is a set of the form or or , where .
(b) Let and be closed disjoint subsets of . For each component of that is an open interval with one endpoint in and the other in , choose a point of . Show that the set of the points is closed.
(c) Show that if is a component of , then does not intersect both and .
- Is every topological group normal?
Exercise after §33
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Examine the proof of the Urysohn lemma, and show that for given ,
rational.
Examine the proof of the Urysohn lemma, and show that for given ,
rational.
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(a) Show that a connected normal space having more than one point is uncountable.
(b) Show that a connected regular space having more than one point is uncountable. [Hint: Any countable space is Lindelöf.]
noteIn fact, there does exist a connected Hausdorff space that is countably infinite; it is called the irrational slope topology.
(a) Show that a connected normal space having more than one point is uncountable.
(b) Show that a connected regular space having more than one point is uncountable. [Hint: Any countable space is Lindelöf.]
In fact, there does exist a connected Hausdorff space that is countably infinite; it is called the irrational slope topology.
- Give a direct proof of the Urysohn lemma for a metric space by setting
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Recall that is a ' set' in if is the intersection of a countable collection of open sets of .
Theorem. Let be normal. There exists a continuous function such that for , and for , if and only if is a closed set in .
A function satisfying the requirements of this theorem is said to vanish precisely on .
Recall that is a ' set' in if is the intersection of a countable collection of open sets of .
Theorem. Let be normal. There exists a continuous function such that for , and for , if and only if is a closed set in .
A function satisfying the requirements of this theorem is said to vanish precisely on .
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Prove:
Theorem (Strong form of the Urysohn lemma). Let be a normal space. There is a continuous function such that for , and for , and otherwise, if and only if and are disjoint closed sets in .
Prove:
Theorem (Strong form of the Urysohn lemma). Let be a normal space. There is a continuous function such that for , and for , and otherwise, if and only if and are disjoint closed sets in .
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A space is said to be perfectly normal if is normal and if every closed set in is a set in .
(a) Show that every metrizable space is perfectly normal.
(b) Show that a perfectly normal space is completely normal. For this reason the condition of perfect normality is sometimes called the ' axiom.' [Hint: Let and be separated sets in . Choose continuous functions that vanish precisely on and respectively. Consider the function .]
(c) There is a familiar space that is completely normal but not perfectly normal. What is it?
A space is said to be perfectly normal if is normal and if every closed set in is a set in .
(a) Show that every metrizable space is perfectly normal.
(b) Show that a perfectly normal space is completely normal. For this reason the condition of perfect normality is sometimes called the ' axiom.' [Hint: Let and be separated sets in . Choose continuous functions that vanish precisely on and respectively. Consider the function .]
(c) There is a familiar space that is completely normal but not perfectly normal. What is it?
- Show that every locally compact Hausdorff space is completely regular.
- Let be completely regular; let and be disjoint closed subsets of . Show that if is compact, there is a continuous function such that and .
Exercise after §34
- Let be a compact Hausdorff space. Show that is metrizable if and only if has a countable basis.
- A space is locally metrizable if each point of has a neighbourhood that is metrizable in the subspace topology. Show that a compact Hausdorff space is metrizable if it is locally metrizable. [Hint: Show that is a finite union of open subspaces, each of which has a countable basis.]
- Show that a regular Lindelöf space is metrizable if it is locally metrizable. [Hint: A closed subspace of a Lindelöf space is Lindelöf.] Regularity is essential; where do you use it in the proof?
Exercise after §35
- Show that the Tietze extension theorem implies the Urysohn lemma.
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Let be a topological space. If is a subspace of , we say that is a retract of if there is a continuous map such that for each .
(a) Show that if is Hausdorff and is a retract of , then is closed in .
(b) Let be a two-point set in . Show that is not a retract of .
(c) Let be the unit circle in ; show that is a retract of , where is the origin. Can you conjecture whether or not is a retract of ?
Let be a topological space. If is a subspace of , we say that is a retract of if there is a continuous map such that for each .
(a) Show that if is Hausdorff and is a retract of , then is closed in .
(b) Let be a two-point set in . Show that is not a retract of .
(c) Let be the unit circle in ; show that is a retract of , where is the origin. Can you conjecture whether or not is a retract of ?
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A space is said to have the universal extension property if for each triple consisting of a normal space , a closed subset of , and a continuous function , there exists an extension of to a continuous map of into .
(a) Show that has the universal extension property.
(b) Show that if is homeomorphic to a retract of , then has the universal extension property.
A space is said to have the universal extension property if for each triple consisting of a normal space , a closed subset of , and a continuous function , there exists an extension of to a continuous map of into .
(a) Show that has the universal extension property.
(b) Show that if is homeomorphic to a retract of , then has the universal extension property.
Exercise after §36
- Prove that every manifold is regular and hence metrizable. Where do you use the Hausdorff condition?
- Let be a compact Hausdorff space. Suppose that for each , there is a neighbourhood of and a positive integer such that can be embedded in . Show that can be embedded in for some positive integer .
- Let be a Hausdorff space such that each point of has a neighbourhood that is homeomorphic with an open subset of . Show that if is compact, then is an -manifold.
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An indexed family of subsets of is said to be a point-finite indexed family if each belongs to for only finitely many values of .
Lemma (The shrinking lemma). Let be a normal space and be a point-finite indexed open covering of , then there exists an indexed open covering of such that for each .
An indexed family of subsets of is said to be a point-finite indexed family if each belongs to for only finitely many values of .
Lemma (The shrinking lemma). Let be a normal space and be a point-finite indexed open covering of , then there exists an indexed open covering of such that for each .
An example of a non-Hausdorff manifold.
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The Hausdorff condition is an essential part of the definition of a manifold; it is not implied by the other parts of the definition. Consider the following space: Let be the union of the set and the two-point set . Topologise by taking as basis the collection of all open intervals in that do not contain , along with all sets of the form and all sets of the form for . The space is called the line with two origins (a.k.a. the bug-eyed line).
(a) Check that this is a basis for a topology.
(b) Show that each of the spaces and is homeomorphic to .
(c) Show that satisfies the axiom, but is not Hausdorff.
(d) Show that satisfies all the conditions for a 1-manifold except for the Hausdorff condition.
An example of a non-Hausdorff manifold.
The Hausdorff condition is an essential part of the definition of a manifold; it is not implied by the other parts of the definition. Consider the following space: Let be the union of the set and the two-point set . Topologise by taking as basis the collection of all open intervals in that do not contain , along with all sets of the form and all sets of the form for . The space is called the line with two origins (a.k.a. the bug-eyed line).
(a) Check that this is a basis for a topology.
(b) Show that each of the spaces and is homeomorphic to .
(c) Show that satisfies the axiom, but is not Hausdorff.
(d) Show that satisfies all the conditions for a 1-manifold except for the Hausdorff condition.