Let be a metric space, and let and be points in such that . By the first axiom of metric spaces, this means that . We can choose with and put and . Then and are open, and and , and we have . This is an obvious and natural construction that occurs frequently in the theory of metric spaces.
Now suppose that is a general topological space, and we again have points with . We might again want to have open sets and with properties as above, but there is no longer any obvious way to produce them, and in fact, they need not exist in all cases. This leads us to introduce the following definition.
Let be a topological space. If with , then a Hausdorff separation for and is a pair of open sets and such that and and . We say that is a Hausdorff space if every pair of distinct points has a Hausdorff separation.
As in our previous discussion, if and are distinct points in a metric space , then the open balls and give a Hausdorff separation of and . Thus, all metric spaces are Hausdorff spaces.
Let be a space with the indiscrete topology, as in Example 3.15, so the only open sets are and . Suppose also that , so we can choose two distinct points with . It is then clear that there can be no Hausdorff separation of and (because both sets and would have to be equal to ). Thus, is not Hausdorff.
The Sierpiński space is defined as follows: the underlying set is , and the sets and are declared to be open, but the set is not open. One can check that this is indeed a topology, and that there is no Hausdorff separation for and , so we do not have a Hausdorff space.
Consider the line with doubled origin as in Example 7.27. This contains two distinct points and . One can check that there is no Hausdorff separation for these points, so is not a Hausdorff space. In particular, this illustrates the fact that a quotient of a Hausdorff space need not be Hausdorff.
Although non-Hausdorff spaces are important in various different branches of mathematics, we will mostly restrict attention to Hausdorff spaces.
Let be a Hausdorff space, and let be a subset of , with the subspace topology. Then is also Hausdorff.
Let and be distinct points in . As is Hausdorff, we can choose sets that are open in with and and . Put and . By the definition of the subspace topology, these are open in . It is also clear that and and , so and give a Hausdorff separation of and in .
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Let and be disjoint topological spaces, and put , and give the coproduct topology as in Definition 7.1. Suppose that and are both Hausdorff; then is also Hausdorff.
Consider a pair of distinct points . There are four possible cases:
Both and lie in
Both and lie in
is in and is in
is in and is in .
In case (a), we use the fact that is Hausdorff. We can thus find sets and that are open in with and and . From the definition of the coproduct topology we see that and are still open when considered as subsets of , so they provide the required Hausdorff separation of and . Case (2) is essentially the same. In case (3) the pair is the required Hausdorff separation, and in case (4) we use instead.
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Let and be topological spaces, and consider the space with the product topology. Suppose that and are both Hausdorff; then is also Hausdorff.
Consider a pair of distinct points and . As , we must either have or . First suppose that in . As is Hausdorff, we can choose a Hausdorff separation for and in . It is then not hard to see that the pair is a Hausdorff separation for and in . Similarly, if then we can choose a Hausdorff separation for and in , and we find that the pair is a Hausdorff separation for and in .
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We now turn to the notion of compactness. Compactness for metric spaces was covered in some detail in [MS, Section 7]. In particular, it was shown as [MS, Theorem 7.8] that a subset is compact iff it is bounded and closed. For our purposes, you should think of this as a prototypical example: a space is compact if it is similar to a bounded, closed subset of . It was also proved in [MS, Theorem 7.20] that a metric space is compact iff it has the Heine-Borel property. In the context of general topological spaces, we take the Heine-Borel property as the definition of compactness. We now recall the relevant details.
Let be a topological space. By an open cover of we mean a family of open subsets such that . Equivalently, for every point there must exist a set in the family such that . By a finite subcover we mean a subcollection , containing only finitely many of the subsets , such that we still have . We say that is compact if every open cover of has a finite subcover.
Consider the space . The sets (for ) form an open cover of . However, we claim that there is no finite subcover. In other words, if we take any finite collection of these sets, say , we claim that the union is not all of . This is clear: if and then and so , for example. From this it follows that is not compact. (We could also have deduced this from the results in [MS, Section 7].)
Now consider the space and the subspaces (for ). These form an open cover of but there is no finite subcover, so is again not compact.
Let be a subset of . We say that is bounded if there is a constant such that for all .
Let be a space with only finitely many points; then is compact.
Supose that . Let be an open cover of . For each , the element must lie in some set of the cover, say . This means that , so we have a finite subcover as required. ∎
Let be a subset of , with the subspace topology. Then is compact iff it is bounded, and closed in .
This follows from [MS, Theorems 7.8 and 7.20]. (It is apparently identical to Theorem 7.8, but that is slightly misleading because [MS] uses a different definition of compactness; we need Theorem 7.20 as well to show that the two versions of compactness are the same.) ∎
The spaces , and are bounded and closed in , so they are all compact.
Recall that . We will show that this is closed and bounded in , so it is compact. We can define by . This is continuous, and is closed in , and , so is closed in . We have also explained that the standard metric on can be expressed in the form
For the identity matrix we have . From this we see that for all , so is bounded.
Let be a compact space, and let be a closed subset of , equipped with the subspace topology. Then is also compact.
Let be an open cover of . By the definition of the subspace topology, we can find sets that are open in such that . As , we see that . Note that the set is also open in (because is assumed to be closed). The set together with the sets then form an open cover of . As is compact, we can choose a finite subcover, which will consist of some finite collection of sets , possibly together with . These sets cover all of , so in particular they cover . It is clear that cannot contribute to covering , so we must have . This means that , so we have a finite subcover of the original cover, as required. ∎
Let be a Hausdorff space, and let be a subset that is compact with respect to the subspace topology. Then is closed in .
We must show that the set is open. Choose a point . For each point we have , so we can choose a Hausdorff separation, say . This means that and are open in and and and . Now put , which is open for the subspace topology on . As , we see that the sets cover . By the compactness property, we can choose a finite subcollection that still covers . This means that . Now put . This is a finite intersection of open sets, so it is still open in . Each of the sets contains , so we have . We also claim that , or equivalently . To see this, suppose (for a contradiction) that we have . This means that , and , so we have for some . On the other hand, we have , so in particular . We now see that , which is impossible because for all . This contradiction shows that we in fact have . We started with an arbitrary element , and we produced an open set that contains and is contained in . By Lemma 3.33, this is enough to prove that is open. ∎
Let be a compact Hausdorff space, and let be a subset of . Then is closed in iff it is compact (in the subspace topology).
Let and be topological spaces. Suppose that is compact, and that we have a surjective, continuous map . Then is also compact.
Consider an open cover of . Put , so that is open by the definition of continuity. If is any point in , then , and the sets cover , so we can choose with , which means that . This shows that the sets form an open cover of . As is assumed to be compact, there must be a finite subcover, say . We claim that the corresponding sets cover . To see this, consider an arbitrary point . As is assumed to be surjective, we can choose with . As the sets cover , we can choose such that . Here was defined to be , so we must have , or in other words as required. Thus, the list is a finite subcover of our original cover. ∎
Let be a compact space, and let be an equivalence relation on . Then the quotient space is also compact.
The quotient map is surjective and continuous, so this follows from Proposition 8.20. ∎
We can define an equivalence relation on by iff . The real projective space is defined to be the quotient space . This is compact, because is compact. Similarly, we can regard as the unit sphere in the space . We can then define another equivalence relation on by declaring that iff for some . The complex projective space is defined to be ; this is again compact.
Let and be disjoint spaces, and take , with the coproduct topology. If and are both compact, then so is .
Let be an open cover of . Then each set must have the form , where is an open subset of , and is an open subset of . It is easy to see that the sets form an open cover of , so there must be a finite subcover, say . Similarly, the sets form an open cover of , so there must be a finite subcover, say . We then find that the list forms a finite subcover of our original cover. ∎
Let and be compact spaces. Then the product is also compact (under the product topology).
We will give the proof after some preliminary definitions and results.
Let and be as above, and let be an open cover of . We will say that a subset is finitely covered if there is a finite subcollection such that .
We need to prove that the whole space is finitely covered, but we will build up to this by proving that certain other sets are finitely covered first.
In the above context, let be a point in . Then there is an open set such that and is finitely covered.
For each , choose an index such that . As is open in the product topology, we can then choose a box such that and and . As for all , we see that the sets form an open cover of . As is compact, we can choose a finite subcover, say . Now put . This is the intersection of a finite list of open sets, each of which contains ; so itself is an open set containing . We next claim that . Indeed, suppose we have a point . As , we can choose such that . Now , so we also have . This proves that , as required. We also have , so
This shows that is finitely covered. ∎
For each , the lemma tells us that we can choose an open set containing such that is finitely covered. The family is then an open cover of . As is assumed to be compact we can choose a finite subcover, say . It follows that . Here each of the subsets is finitely covered, and there are only finitely many of them, so the union is also finitely covered. This means that is finitely covered, or in other words that our original cover of has a finite subcover, as required. ∎
Let be a compact space, and let be a Hausdorff space. Let be a continuous bijection. Then the inverse map is also continuous, so is a homeomorphism.
Let be the inverse of , so the claim is that is continuous. By Proposition 3.23, it will be enough to check that for every closed subset , the preimage is closed in . Here is a closed subset of a compact space, so it is compact by Proposition 8.17. As is inverse to , we have . We can regard as a continuous surjective map from to (where and are given the respective subspace topologies). It follows by Proposition 8.20 that is also compact. Now is a compact subspace of the Hausdorff space , so it is closed by Proposition 8.18. In other words, the preimage is closed, as required. ∎
In Example 7.24 we considered a space consisting of two disjoint discs, and the space . We introduced an equivalence relation on (corresponding to the idea of gluing the boundary circles of the two discs) and defined a continuous bijection . Now is a quotient of a closed bounded subset of , so it is compact, and is a metric space, so it is Hausdorff. It follows from Proposition 8.27 that is actually a homeomorphism.
We conclude with some slightly different results about open covers, that are only relevant for metric spaces. Their use will become apparent later.
Let be a metric space, and let be an open cover. A Lebesgue number for this cover is a number with the following property: for every point , there is an index such that .
You should think of a Lebesgue number as something like the minimum size of overlaps between adjacent sets in the cover.
Let be a compact metric space. Then every open cover of has a Lebesgue number.
Let be an open cover of . For each point , we can choose an index such that . As is open, we can then choose such that . Put , so is an open subset of containing . The sets then form an open cover of , so we can choose a finite subcover, say . Put . We claim that this is a Lebesgue number. To see this, let be an arbitrary point in . As the sets form a cover, we can choose such that , or in other words . If then and so . This proves that , so is contained in one of the sets , as required. ∎