User:TakuyaMurata/Topological groups

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Topological spaces[edit]

Given an arbitrary set , a subset of the power set of is called a topology for if it includes

  • (i) the empty set and
  • (ii) the union of any subset of , and
  • (iii) the intersection of any two members of .

The members of are said to be open in . Immediate examples of topologies are and the power set of . They are called trivial and discrete topologies, respectively. In the discrete topology, every set is open. On the other hand, and are the only subsets that are open in the trivial topology.

In application, we are usually given some set (which often carries an intrinsic structure on its own) and then induce a topology to by defining a according to how we want to do analysis on . The key insight here is that saying some set is open and another closed is merely the matter of labeling. One way to induce a topology is to combine existing ones. For that, we use:

1 Lemma The intersection of topologies for the same set is again a topology for the set.
Proof: Let be a family of topologies for the same set. If , then every member of , which is closed under unions. Thus, is in every member of . The other properties can be verified in the same manner.

Another common method of inducing a topology is to define topology from convergence of filters. A filter is said to converge to if contains every neighborhood of . The convergence depends on which sets are considered neighborhoods of ; in other words, it depends on a topology. We could go in the other way: given a collection of filters , we declare sets to be open so that . We only have to check this in fact defines a topology.

Let be a function to some topological space. Then the set of for every open set is a the weakest among topologies that make continuous. If is a family of functions defined on the same set, then a weak topology generated by is the intersection of the weakest topology that makes each member of continuous. A weak topology is indeed a topology since the intersection of topologies for the same set is again a topology by Lemma.

Two important types of topological spaces are constructed by this method: product and quotient spaces.

A subset is said closed in when is open. The intersection of all closed set containing is called the closure of and denoted by . By (iii) in the definition, the closure of is closed. Moreover, a set is closed if and only if .

1 Lemma

  • (i) where the inequality holds if is finite, and
  • (ii) .

Proof: (i) Since for all the desired inequality holds, and since the finite union of closed sets is closed the equality holds if is finite. A similar argument shows (ii).

A neighborhood of a point is an "open" subset of containing . A point is in the closure of if and only if every neighborhood of intersects . Paraphrasing in the language of filter, is in the closure of if and only if a filter containing converges to .

In a sense, a filter is a generalization of the notion of "neighborhoods of a point." This explains why a filter, by definition, doesn't contain the empty set (the empty set is a neighborhood of no point as well as the property is in a filter then any set containing is in the filter (if is a neighborhood of a point , then any larger set should also be a neighborhood of that point.) In fact, given a point in some set , let . Then is an ultrafilter on .

Let . We say is continuous when its pre-image of every open set in is open.

1.4 Theorem Let . The following are equivalent:

  • (i) is continuous.
  • (ii) If is an open subset and , then there exists a neighborhood of such that .
  • (iii) If a filter , then .

Proof: (i) (ii) means that . Thus, is a neighborhood of and . Take . (ii) (iii): If is a neighborhood of , then there is as in (ii). Since is a neighborhood of , by convergence, ; thus, and so . (iii) (i): We claim:

for any subset . Suppose a filter containing converges to some point . By (iii), . This proves the claim. (By the way, the claim is then another equivalent definition of continuity.) Now, given a closed subset , applying the claim we get:

Thus, and so is closed. Since , is an open mapping then.

In particular, (iii) means that if is a discrete space, then every function on is continuous.

1 Theorem The composite of continuous functions is again continuous.
Proof: If a filter , then and so .

A subset of a topological space is locally closed if it can be written as the intersection of a closed set and open set. (Note: sets here are allowed to be empty; so, closed sets and open sets are locally closed.) Equivalently, a set is locally closed if and only if it is open in its closure.

A topological space is said to be connected if it is not a disjoint union of nonempty open sets. Equivalently, a topological space is connected if it has no proper open closed subsets.

A topological space is said to be irreducible if it cannot be written as a union of two proper closed subsets.

1. Theorem Let X be a topological space. Then the following are equivalent.

  • (i) X is irreducible.
  • (ii) Every nonempty open subset of X is dense.
  • (iii) Every open subset of X (in particular X) is connected.

Proof: (i) (ii): Let be a nonempty open subset. X is then the union of and the complement of U. By (i), cannot be a proper subset. (ii) (iii): Let A and B be open subsets. By (ii), is dense and so intersects B. (iii) (i): If X is not irreducible, then X is the union of proper closed subsets A and B. Then is the disjoint union of nonempty open subsets; thus, not connected.

If X is irreducible, then is a domain. (Consider the zero set of the product )

1. Corollary A continuous image of an irreducible space X is again irreducible.
Proof: Let be a continuous map. Let be an open subset. Then is open. By (ii) in the theorem and continuity, we have:

.

Compact sets and Hausdorff spaces[edit]

A subset of a topological space is said to be compact if every ultrafilter in it converges to exactly one point. (Note: This definition, due to Bourbaki, is slightly different but much simpler than one in literature.) In particular, a compact set is closed, and a closed subset of a compact space is compact.

1 Theorem A continuous function sends compact sets to compact sets.
Proof: Let be a compact set, and suppose is an ultrafilter on K. Then s is an ultrafilter and so converges to, say, x. By continuity, f[s] converges to f(x).

1 Corollary A function with compact graph is continuous.
Proof: Let be a function that has compact graph. Let and be canonical surjection and injection, respectively. Then . By hypothesis, is a closed map; its inverse is thus continuous. Hence, is continuous.

A function is said to be proper if the pre-image of a compact set is compact.

1 Theorem A function is proper if and only if

implies .

Here means that every compact set contains only finite many .
Proof: () is obvious. For the converse, suppose f is not proper. Then there exists a compact subset such that is not compact. The non-compactness allows us to find a strictly increasing sequence of open subsets such that for every but the union contains . Inductively, we can then find a sequence such that . Now, since for all . On the other hand, . Indeed, suppose is a compact subset. Since consists of isolated points; thus, closed, the set is a compact subset, of which is an open cover. Thus, there is some such that . Thus, for all .

1 Theorem Every proper map into a locally compact space is a closed map. (Recall that we assume every locally compact space is Hausdorff.)

1 Theorem Every continuous bijection from a compact space to a Hausdorff space is a homeomorphism. 1 Theorem A topological space X is compact if and only if, for all topological spaces Y, the projection is closed.

A Hausdorff space X is said to be compactly generated if closed subsets A of X are exactly subsets such that is closed in K for all compact subsets .

1 Theorem (Tychonoff's product theorem) The following are equivanelt:

  • (i) Every product space of compact spaces is compact.
  • (ii) Axiom of Choice.

Proof: (i) (ii). Let be a collection of compact spaces and be a projection from . Let be an ultrafilter on . For each , since is again an ultrafilter and is compact, must converge. Then it follows that converges. Axiom of Choice implies that the product space is compact. Conversely, let be a nonempty collection of nonempty sets. Let be a point such that for all . Such a must exist; if not, the intersection of is the universal set, contradicting that it is a proper class. For each , let and . Then is a topology for and since finiteness is compact. Using (i) is compact and thus has the finite intersection property and this implies the statement equivalent to (ii).

A topological space X is called a Tychonoff space if it is Hausdorff and, given a closed set F and a point x outside F, there exists a continuous function such that on F and .

1 Theorem Every locally compact space is Tychonoff.

1 Theorem (Stone-Čech compactification) Let X be a Tychonoff space. Then there exists a compact space and a continuous injection such that for any continuous map (where is a compact space) there is a unique continuous map with

Stone–Cech compactification.png

Proof: See [1] (See also: [2])


A cover of the set E is a collection of sets {} such that . An open cover of is a cover of consisting of open sets, or equivalently, a subset of the topology whose union contains . A subset of cover of is called a subcover if it is again a cover of .

1 Theorem A topological space is compact if and only if

  • (i) Every open cover of contains a subcover that is a finite set, and
  • (ii) X is Hausdorff.

Proof: () (ii) is immediate. For (i), we first remark that the following are equivalent:

  • (a) Every nonempty collection of closed subsets of with the finite intersection property is nonempty.
  • (b) Every nonempty collection of closed subsets of with empty intersection contains a finite subcollection with empty intersection.
  • (i).

To see the equivalence of (i) and (b), consider the collection consisting of , given an open cover . We shall prove (a). Let be a nonempty collection of closed sets with the finite intersection property, and be an ultrafilter containing . Since converges to, say, ,

.

() Because of (ii), we only have to show that every ultrafilter converges to some point. Let be an ultrafilter. Since has the finite intersection property, by an equivalent form of (i), has a point, say, . For every neighborhood of , we have either or . The latter not being the case, we have . In other words,

1 Corollary If is a sequence of compact sets, then is nonempty.

A set is compact if every open cover {} of it has a finite subcover; i.e.,

for some .

For example, let be open cover of the finite set . Then for each , we can find some . It thus follows that a finite set (e.g., the empty set) is compact since

Conversely, every compact subset of a discrete space is finite.

1 Theorem If a topological space is the union of countably many compact sets, then any of its open cover admits a countable subcover.
Proof: Let be a sequence of compact sets, and suppose that its union is covered by some open cover . For each , since is an open cover of , it admits a finite subcover of . Now, is a countable subcover of . .

1 Theorem The following are equivalent:

  • (i) Axiom of Choice.
  • (ii) Every product topology of compact topologies is compact. (Tychonoff's product theorem)
  • (iii) Something has empty intersection.

Proof: Let be a product topology and be a collection of projections on . Let be an ultrafilter on . For each , since is again an ultrafilter and is compact, converges. From the lemma 1.something it follows that converges. If (i) is true, then the convergence implies that is compact. To show (ii) implies (iii), Let be a nonempty collection of nonempty sets. Also, let be a element such that . Such an element must exist since the contrary means that is the universal set. For each let and induce the topology by letting . Then since its topology is finite, each is compact. That (iii) implies (i) is well known in set theory.

We say a point is isolated if the set is both open and closed, otherwise called an limit point. If a set has no limit point, then the set is said to be discrete.

1 Lemma Every infinite subset of a compact space has a limit point.
Proof: Let be infinite and discrete. Then is closed since it contains all of limit points of . Since is a closed subset of a compact set, is compact. It now follows: for each , the singleton is open and thus the collection is an open cover of , which admits a finite subcover. Hence, we have:

,

contradicting that is infinite. .

The points x and y, separated by their respective neighbourhoods U and V.

We say a topological space is Hausdorff if two distinct points are covered by disjoint open sets. This definition can be strengthened considerably.

1 Theorem A topological space is Hausdorff if and only if every pair of disjoint compact subsets of can be covered by two disjoint open sets.
Proof: Let be compact and disjoint, and be fixed. For each we can find disjoint open sets and such that and . Since is compact, there is a finite sequence such that:

.

Let , and . Since is disjoint from any of , and are disjoint. is open since it is a finite intersection of open sets. Since is compact, there is a finite sequence such that:

.

1 Theorem A topological space is Hausdorff if and only if a filter on it converges, if it ever does, to at most one point.
Proof: Suppose converges to two distinct points and . By the separation axioms, we can find disjoint subjects and such that and . Since convergence, . But this then implies that , a contradiction.

A Hausdorff space is said to be locally compact if every point in it has a compact neighborhood. A locally compact space is thus locally closed.

We also give two other characterizations of Hausdorff spaces, which are sometimes useful in application.

1 Lemma A topological space is Hausdorff if and only if for every point the set is the intersection of all of its closed neighborhoods.

1 Lemma Let be a family of functions from to a Hausdorff space . Let be the union of taken all over open sets of and . Then satisfies the Hausdorff separation axiom if and only if for each with , there is some such that .
Proof: First suppose the separation axiom. Then we can find two disjoint sets such that and . Then since by definition, there is some function and sets and such that and . Thus, . Conversely, suppose the family separates points in . Then by the separation axiom there are disjoint open open sets and such that and . Then by the definition of and are disjoint and both open.

1 Theorem Let be continuous. If is Hausdorff, then the graph of is closed.
Proof: Let be the complement of the graph of . If , then, since and is Hausdorff, we can find in disjoint neighborhoods and of and , respectively. It follows: since there is no point such that and . By continuity, is open; thus, is open.

1 Corollary Let be continuous functions. Suppose is Hausdorff. If on some dense subset, then identically.
Proof: Let be the intersection of the graph of and the graph of . is dense in the graph of by hypothesis and closed by the preceding theorem.

1 Theorem A dense subspace X of a compact Hausdorff space Y is locally compact if and only if X is an open subset of Y.

A graph of a function is the set consisting of ordered pairs for all the domain of . In the set-theoretic view, of course this set is . But since we usually do not see a function as a set, the notion is often handy to use.

1 Lemma Let be a function. Suppose is locally compact. Then is continuous if and only if is continuos on every compact subset of .
Proof: Suppose where is an open subset. By local compactness, has a neighborhood where is a compact set. Since is continuous by hypothesis, has a neighborhood such that: .

Topological groups[edit]

Let G be a topological group, which we don't assume to be Hausdorff.

1 Theorem

  • (i) G is Hausdorff if and only if is closed.
  • (ii) If G is Hausdorff, then its discrete subgroup is closed.

Proof: Left to the reader.

1 Theorem (Baire) Let be a locally compact space. If is written as a union of countably many sets, then one of those sets contains a nonempty open subset.
Proof: Similar to one given in w:Baire category theorem.

1 Corollary If is countable and locally compact, then is discrete.
Proof: We write . Then is then open for some . It follows that and so is open. (Recall that every finite subset of a Hausdorff space is closed.) Hence, every subset of is open and closed.

In particular, the only topology that makes a locally compact topological group is the discrete one.

1. Theorem Let be a compact subset. Then consists of uniformly continuous functions.
Proof: We only show right uniform continuity, since the proof for the left uniform continuity is completely analogous. Let , and be given. By continuity, for each , there is a neighborhood of e such that

    for all .

(Note: by the continuity of translation, for any neighborhood V of e, one can always find another neighborhood W of e such that .) By compactness, is contained in the union of some . Let . It follows that:

    for all .

Indeed, if , then for some and so:

    for all .

for any .

1. Corollay Every function in is uniformly continuous. (Actually, true for C_0(G)?)

References[edit]

Notes[edit]

Most of materials in "Compact sets and Hausdorff space" section comes from [3]