We will now change direction, and spend some time building the required theory of topological spaces, which will serve as a foundation for the rigorous definition of homology groups.
Topological spaces are a generalisation of metric spaces. Knowledge of metric spaces is mostly assumed as a prerequisite, but will be reviewed briefly here. References marked [MS] refer to Dr Roxanas’s notes for MAS331 (Metric Spaces), but equivalent results can be found in many other sources.
Let be a set. A metric on is a function with properties as follows:
For all we have , and iff .
For all we have .
For all we have (the Triangle Inequality).
A metric space is a set equipped with a metric.
We can define three different metrics on , as follows:
Later we will explain a sense in which these are essentially the same.
Consider the set of real matrices (or any of the various subsets of discussed in Example 1.4). We can define a metric as follows:
This is not really a new example. A little matrix algebra shows that , so if we identify with in the obvious way, then this new metric is just the same as the standard metric in the previous example. However, this new formula for the metric makes it easier to combine with other constructions in matrix theory, such as the definition of the orthogonal group.
We will not need any examples that are much more exotic than these.
Let and be metric spaces, with metrics and . Let be a function from to . We say that is continuous if it has the following property:
For all and , there exists such that for all with , we have .
Suppose that satisfies for all . Then is continuous.
Suppose we are given and . We need to provide a number and check that it has a certain property. We just take . Suppose that we have with . By our assumptions on , and , we then have . This is the required property of . ∎
Let be a metric space, let be a point of , and let be a positive real number. We put
and call this the open ball of radius centred at .
Let be a metric space, and let be a subset of . We say that is open if for every point , there exists such that .
We also say that a subset is closed if the complement is open.
Our definition of closed sets is not the same as in [MS, page 40], but it is equivalent to that definition, as proved in [MS, page 43].
In any metric space :
The empty set and the whole set are open.
The union of any collection of open sets is open.
The intersection of any finite collection of open sets is open.
One of the main reasons for introducing metric spaces is to formalise the notion of a continuous map. The following result shows that we can do that using only the system of open sets, we do not actually need the metric.
Let be any map of sets, and let be a subset of . We put
and we call this the preimage of under .
Let be any map of sets.
For any family of subsets , we have and .
For any subset we have .
If is another map of sets, and , then we have .
For the first claim, we have iff iff ( for all ) iff ( for all ) iff . All the other claims can be proved in a similar way. ∎
Let be a function between metric spaces. Then the following are equivalent:
is continuous.
For every open set , the preimage is an open subset of .
For every closed set , the preimage is a closed subset of .
We can now introduce the theory of topological spaces, which was mentioned briefly at the end of [MS, Section 3].
Let be a set. A topology on is a collection of subsets of (which are called open sets) with the following properties:
The empty set and the whole set are open.
The union of any collection of open sets is open.
The intersection of any finite collection of open sets is open.
For any set , we can introduce a topology by declaring that every subset is open. This is called the discrete topology. Although this is not very interesting, it does occur naturally: it is the most natural topology on for example. At the other extreme, we can introduce a different topology by declaring that only the sets and are open. This is called the indiscrete topology. Unlike the discrete topology, this is rarely relevant.
Suppose we have two different metrics on , say and . Suppose we also have positive constants and such that and . Then a set is open with respect to iff it is open with respect to . Thus, the metrics and give the same topology.
Suppose that is open with respect to . Consider a point . By assumption, there exists such that , where is the open ball defined using , or in other words
We claim that is also contained in . Indeed, if , then , so , so . However, we have by assumption, so . This proves that as claimed. As we can do this for any , we see that is open with respect to . By a symmetrical argument, if is any set that is open with respect to , then it is also open with respect to . Thus and have the same open sets, and thus the same topology. ∎
The metrics , and all give the same topology on .
In the light of the lemma, it will be enough to prove the inequalities
Put , then choose such that is the largest of all the terms . We then have
so the required inequalities are
For the third inequality, the sum contains terms each of which is less than or equal to , so the total is at most , as required. The other two inequalities are equivalent to
The first sum consists of plus some other nonnegative terms. In the second sum, the terms with give the same as the first sum, and there are again some additional nonnegative terms. The claim is clear from this. ∎
For put
The inequalities in the above proof are equivalent to the claim that . This can be illustrated in the case as follows:
Let and be topological spaces, and let be a function from to . We say that is continuous if for every open set , the preimage is open in .
Proposition 3.12 shows that when and are metric spaces with the metric topology, this agrees with our earlier definition of continuity.
We can define a function by
The functions , and are known to be continuous, by basic real analysis. It is also standard that sums and products of continuous functions are continuous, and it follows that each of the four components of is continuous. From this it follows easily that itself is continuous. In future, we will not bother to discuss this kind of argument in detail.
As a very basic example, consider a constant function . This means that there is a single point such that for all . Consider an open set . If then , and if then . As both and are open in , we see that is always open. This proves that is continuous.
Let be as above. Then is continuous iff for every closed subset , the preimage is closed in .
Suppose that is continuous. Let be closed; we must show that is closed in . As is closed in , we know that the complement must be open in , so is open in (by the definition of continuity). However, is the same as . As this is open, we see that must be closed, as required.
The converse can be proved in essentially the same way. ∎
Suppose that , and are topological spaces, and we have continuous maps and . Then the composite map is also continuous. Moreover, the identity map is also continuous.
Consider an open set ; we must show that the preimage is open in . As is continuous, we see that is an open subset of . As is continuous, it follows in turn that is open in . However, we have iff iff iff , so is the same as . We have therefore proved that is open in , as required.
For the second claim, suppose that is an open subset of . Then is just the same as , and so is open. This proves that is continuous. ∎
Let be a topological space, and let be a subset of . We declare that a subset is open in if there exists an open set of such that .
The above definition gives a topology on (which we call the subspace topology).
The empty set can be written as the intersection of with the open set of , so the empty set is open in . The full set can be written as the intersection of with the open set of , so is also open in .
Suppose we have a collection of sets that are open in ; we must show that the union is also open in . As each is open in , we can find open sets such that . We are assuming that the axioms for a topology are satisfied by the open sets in , so the union is again open in . The set can be written as , so it is open as required.
Now suppose instead that we have a finite collection of sets that are open in ; we must show that the intersection is also open in . As in (b), we can choose open subsets with . We are assuming that the axioms for a topology are satisfied by the open sets in , so the intersection is again open in . The set can be written as , so it is open as required.
∎
One very basic point about the subspace topology is as follows. We can define by for all . We call this the inclusion map.
If we give the subspace topology, then the inclusion map is continuous.
Consider an open set ; we must show that is open with respect to the subspace topology on . But is just the same as , which is open by the definition of the subspace topology. ∎
We next deal with a slightly technical point. Suppose, for example, we want to discuss whether the function is continuous. We might want to regard this as a function from to , or as a function from to . We also might want to restrict the range of values of , and consider as a function defined on or or instead of all of . This leads us to worry about the following possibility: perhaps some of these versions of are continuous, and some of them are not. It would then be a nightmare to keep track of everything. Fortunately, however, this nightmare does not arise: if we can check that is continuous as a map , then all other versions of are automatically continuous. This is the message of the following proposition.
Let and be topological spaces. Let be a subset of , and let be a subset of , with inclusion maps and . Let be a continuous function from to . Let be the corresponding map , obtained by restricting the domain to and enlarging the codomain to , so we have a commutative diagram as follows:
Then is also continuous.
There is also a partial converse for the above result.
Let and be topological spaces, and let be the inclusion of a subset with the subspace topology. Let be a function from to , and suppose that is continuous (or equivalently: is continuous when regarded as a function ). Then is continuous (as a function ).
Let be a subset of that is open with respect to the subspace topology. We must show that is open in . By the definition of the subspace topology, there must exist an open set such that . This can also be written as , so . We are given that is continuous and is open in so is open in . In other words, is open in as required. ∎
Let be an open subset of , and let be a subset of . Then is open with respect to the subspace topology on iff is open with respect to the original topology on .
First suppose that is open with respect to the subspace topology. By definition, this means that for some subset that is open in . Now both and are open in , so the intersection is also open in , as required.
Suppose instead that we start from the assumption that is open with respect to the original topology on . If we just take , we see that is open in and , so is open with respect to the subspace topology. ∎
Let be an arbitrary subset of , and let be a subset of . Then is closed (with respect to the subspace topology on ) iff there exists a closed set such that .
If is closed in , then the complement must be open in . This means by definition that there exists an open set such that . We can now put , which is a closed subset of . We find that
so has the form , as required. We leave the converse to the reader. ∎
Let be a closed subset of , and let be a subset of . Then is closed with respect to the subspace topology on iff is closed with respect to the original topology on .
Let be a topological space, and let be a subset of . Suppose that for each we can find an open set such that and . Then itself is open.
For each point , choose a set as described, so is open and and . Put . This is the union of a family of open sets, so it is open by the axioms for a topology. If we can prove that is the same as , then we will be done. Each set is contained in , and it follows that the union is also contained in . On the other hand, for each we have , so . As and we have as required. ∎
Let be a function between topological spaces. Suppose we have subsets such that
Each set is open.
For each , the restricted map is continuous (with respect to the subspace topology on ).
Then is continuous.
Let be an open subset of ; we must show that is open in . Let be the restriction of . As is continuous by assumption, we know that is open in . By Lemma 3.30, this is the same as being open in . Moreover, we see from the definitions that . As , we see that
We have seen that each set is open, so the union of these sets is open, so is open as required. ∎
Again let be a function between topological spaces. Suppose we have subsets such that
Each set is closed.
For each , the restricted map is continuous (with respect to the subspace topology on ).
Then is continuous.
By Proposition 3.23, it will be enough to show that is closed in whenever is closed in . After this preliminary step, the rest of the proof is essentially the same as for the previous proposition. ∎