Lecture 6: September 15 Connected Components. If a Topological Space X Is Not Connected, It Makes Sense to Investigate the Maximal Connected Subspaces It Contains

Home , Connected space

Lecture 6: September 15 Connected components. If a topological space X is not connected, it makes sense to investigate the maximal connected subspaces it contains. Given a point x X,wedefine ∈ C(x)= C X C is connected and x C ⊆ ∈ as the union of all connected subspaces of X that contain the point x. Proposition 6.1. These sets have the following properties: (a) Each subspace C(x) is connected and closed. (b) If x, y X, then C(x) and C(y) are either equal or disjoint. (c) Every nonempty∈ connected subspace of X is contained in a unique C(x). Proof. C(x) is a union of connected subspaces that all contain the point x, and therefore connected by Proposition 5.8. Now Proposition 5.9 shows that the closure C(x) is also connected; since it contains x, it must be contained in C(x), and therefore equal to C(x). This proves (a). To prove (b), suppose that C(x) and C(y) are not disjoint. Take any z C(x) C(y). We shall argue that C(x)=C(z); by symmetry, this will imply that∈ C(x)=∩ C(z)=C(y). We have z C(x), and since C(x) is connected, we get C(x) C(z). Hence x C(z), and for∈ the same reason, C(z) C(x). To⊆ prove (c), let Y ∈ X be nonempty connected subspace.⊆ Choose a point x Y ;thenY C(x) by⊆ construction. Uniqueness follows from (b). ∈ ⊆ The set C(x) is called the connected component of x; the proposition shows that it is the maximal connected subspace of X containing the point x. Since any two connected components of X are either equal or disjoint, we get a partition of the space X into maximal connected subsets. Example 6.2. The connected components of a space are always closed, but not necessarily open. In the case of Q, the connected component of x Q is x , because no subspace with two or more points is connected. So the∈ connected{ } components of X do not give us a separation of X in general. Example 6.3. If X has only finitely many connected components, then each connected component is open (being the complement of a finite union of closed sets). There is a similar definition for path connectedness: given a point x X,weset ∈ P (x)= P X P is path connected and x P ⊆ ∈ = y X x and y can be joined by a path inX , ∈ and call it the path component of x. The following result can be proved in the same way as Proposition 6.1. Proposition 6.4. The path components of X have the following properties: (a) Each subspace P (x) is path connected. (b) If x, y X, then P (x) and P (y) are either equal or disjoint. (c) Every∈ nonempty path connected subspace of X is contained in a unique P (x). Since P (x) is connected by Proposition 5.14, it is contained in C(x); but the two need not be equal. 2

Example 6.5. The topologist’s sine curve is connected, but it has two path components. Path connectedness and connectedness. I mentioned last time that in suﬃ- ciently nice topological spaces (such as open subsets of Rn) connectedness and path connectedness are equivalent. The point is that Rn contains lots and lots of paths, whereas a random topological space may not contain any nonconstant path at all. This suggests looking at spaces where at least any two nearby points can be joined by a path. Deﬁnition 6.6. A topological space X is called locally path connected if every point has a path connected neighborhood.

Open balls in Rn are obviously path connected. Consequently, every open set in Rn is locally path connected; more generally, every topological manifold is locally path connected. Proposition 6.7. Let X be a locally path connected topological space. Then P (x) is open, and P (x)=C(x) for every x X. In particular, the connected components of X are open. ∈ Proof. You will remember a similar result from one of last week’s homework prob- lems. We ﬁrst prove that P (x) is open for every x X.SinceX is locally path connected, there is a path connected neighborhood U∈of x;wehaveU P (x), because P (x) is the union of all path connected subspaces containing x.If⊆ y P (x) is any point, then P (y)=P (x), and so P (x) also contains a neighborhood∈ of y, proving that P (x) is open. Next, we show that P (x)=C(x). The union of all the other path components of X is open, and so their complement P (x) must be closed. Now P (x) C(x)isa nonempty subset that is both open and closed; because C(x) is connected,⊆ it must be that P (x)=C(x). The last step in the proof is a typical application of connectedness: to prove that a nonempty subset Y of a connected space X is equal to X, it is enough to show that Y is both open and closed.

Compactness. The second important property of closed intervals in R is compactness. Perhaps you know the definition that is used in analysis: a subset A Rn is called compact if every sequence in A has a convergent subsequence. As⊆ with closed sets, definitions involving sequences are not suitable for general topological spaces. We therefore adopt the following definition in terms of open coverings. Definition 6.8. An open covering of a topological space X is a collection U of open subsets with the property that X = U. U U ∈ We say that X is compact if, for every open covering U , there are finitely many open sets U ,...,U U such that X = U U . 1 n ∈ 1 ∪···∪ n Compactness is a very important finiteness property of the topology on a space. In fact, every topological space with only finitely many points is trivially compact; in a sense, compact spaces are thus a generalization of finite topological spaces. 3

Note that compactness is a topological property: if X is compact, then any space homeomorphic to X is also compact. You should convince yourself that when the topology of X is given in terms of a basis, it suﬃces to check the condition for open coverings by basic open sets. Example 6.9. R is not compact: in the open covering

∞ R = ( n, n), − n=1 no finite number of subsets is enough to cover R. Example 6.10. Q [0, 1] is also not compact. This is obvious using the analyst’s definition in terms∩ of sequences; here is one way of proving it with our definition. We enumerate all the rational numbers between 0 and 1 in the form α1,α2,..., and also choose an irrational number β [0, 1]. For every k,letIk be the open interval of length = α β centered at∈ the point α . Obviously, the sets k | k − | k Ik Q [0, 1] ∩ ∩ form an open covering of Q [0, 1]; I claim that no finite number of sets can cover everything. Otherwise, there∩ would be some integer n 1 such that every rational number between 0 and 1 lies in I I . But then≥ the open interval of length 1 ∪···∪ n min k 1 k n ≤ ≤ centered at β would not contain any rational number, which is absurd. These examples illustrate an important point: it is usually pretty easy to show that a space is not compact, because all we have to do is find one offending open covering. On the other hand, it can be quite hard to show that a space is compact, because we need to consider all possible open coverings. Example 6.11. The closed unit interval [0, 1] is compact; we will prove this next time. General properties of compactness. The following lemma is an easy conse- quence of the definitions (of compactness and of the subspace topology). Lemma 6.12. Let X be a topological space, and let Y X be a subspace. The following two conditions are equivalent: ⊆ (a) Y is compact (in the subspace topology). (b) Whenever U is a collection of open sets in X with Y U, ⊆ U U ∈ there are finitely many sets U ,...,U U with Y U U . 1 n ∈ ⊆ 1 ∪···∪ n We can use this lemma to prove some general properties of compact spaces. Proposition 6.13. Every closed subspace of a compact space is compact. Proof. Let X be compact and Y X closed. Given an open covering U of Y ,we get an open covering of X by throwing⊆ in the open subset U = X Y .SinceX 0 \ is compact, there are finitely many sets U1,...,Un U such that X = U0 U1 U .ButthenY U U , proving that ∈Y is compact. ∪ ∪ ···∪ n ⊆ 1 ∪···∪ n 4

Proposition 6.14. If X is compact and f : X Y is continuous, then f(X) is again compact. → Proof. Let U be an arbitrary open covering of f(X). Then 1 f − (U) U U ∈ is a collection of open sets that cover X; because X is compact, there must be ﬁnitely many sets U ,...,U U with 1 n ∈ 1 1 X = f − (U ) f − (U ). 1 ∪···∪ n Since U ,...,U f(X), we now get f(X)=U U , proving that f(X)is 1 n ⊆ 1 ∪···∪ n also compact.