A separation of a topological space \(X\) is a pair of disjoint open sets \(U_1, U_2\) satisfying \(X=U_1\cup U_2\text{.}\) Call a separation \(X=U_1\cup U_2\) trivial if \(\{U_1, U_2\}=\{\emptyset, X\}\text{,}\) and nontrivial otherwise. A space \(X\) is connected if it is nonempty and all separations of \(X\) are trivial; it is disconnected otherwise.
A subset \(Y\subseteq X\) is connected (resp., disconnected) if it is connected (resp., disconnected) with respect to the subspace topology.
Remark1.12.2.Conventions.
Our defined terms in Definition 1.12.1 differ from Munkres in two regards.
Munkres takes as a separation what we call a nontrivial separation. We choose the weaker notion so that we have a convenient term for a disjoint cover of a space by two open sets.
Munkres does not insist that a space be nonempty to be connected. There does not seem to be much consensus on whether to include this as a necessary condition. However, when discussing connected components later on, we would like to be able to say simply that a space is connected if and only if it has exactly one connected component. This is reason enough for us to go with the nonempty condition.
Example1.12.3.Examples.
If \(\abs{X}\geq 2\text{,}\) then \(X\) is not connected with respect to the discrete topology.
If \(X\) is infinite, then \(X\) is connected with respect to the cofinite topology.
Let \(L\) be the graph of the equation \(y=0\) and let \(C\) be the graph of the equation \(y=1/(1+x^2)\text{.}\) The subset \(X=L\cup C\subseteq \R^2\) is disconnected.
\(Q\subseteq \R\) is not connected. In fact if \(A\subseteq \R\) is any set that is not an interval, then \(A\) is not connected. (Define an interval of \(\R\) to be a set \(I\) satisfying the following property: if \(a,b\in I\) and \(a < b\text{,}\) then \([a,b]\subseteq I\text{.}\) This definition includes singletons and the empty set as sort of degenerate intervals.)
Solution.
Theorem1.12.4.Connected subsets of \(\R\).
Consider \(\R\) with the standard topology. The connected subsets of \(\R\) are precisely the nonempty intervals of \(\R\) (including “infinite intervals” like \((a,\infty)\) or \([b,\infty)\)): i.e., \(Y\subseteq \R\) is connected if and only if \(Y\) is a nonempty interval.
More generally, if \(X\) is a space equipped with an order topology, then a subset \(Y\subseteq X\) is connected if and only if
\(Y\) satisfies the least upper bound principle (i.e., every bounded subset of \(Y\) has a supremum in \(Y\)), and
if \(x,y\in Y\) and \(x< y\text{,}\) then there exists \(z\in Y\) satisfying \(x< z< y\text{.}\)
Proof.
See Munkres.
Theorem1.12.5.Connectedness equivalences.
The following statements are equivalent for a topological space \(X\text{.}\)
\(X\) is connected.
There is no pair of disjoint nonempty closed sets \(C_1, C_2\) satisfying \(X=C_1\cup C_2\text{.}\)
If \(U\subseteq X\) is open and closed, then \(U\in \{\emptyset, X\}\text{.}\)
There is no pair of nonempty disjoint sets \(A, B\text{,}\) neither of which contains limit points of the other, satisfying \(X=A\cup B\text{.}\)
Proof.
\((1)\iff (2) \iff (3)\text{.}\) It is clear that a separation \(X=U\cup V\) by open sets is the same thing as a separation as described in (2), since \(U=V^c\) and \(V=U^c\) are both open and closed. Similarly, we see that there is a separation \(X=U\cup V\) if and only if and only if there is a nonempty open set \(U\) whose complement \(U^c\) is open and nonempty, if and only if there is a nontrival \(U\) that is open and closed.
\((2)\iff (4)\text{.}\) This equivalence follows from the observation that we have \(X=A\cup B\) with \(A, B\) as described in (4) if and only if \(A=\overline{A}\) and \(B=\overline{B}\text{,}\) if and only if \(A\) and \(B\) are disjoint and closed.
Theorem1.12.6.Connectedness and subspaces.
Let \(X\) be a topological space.
If \(Y\) is a connected subspace of \(X\text{,}\) and if \(X=U_1\cup U_2\) is a separation of \(X\text{,}\) then \(Y\subseteq U_1\) or \(Y\subseteq U_2\text{.}\)
If \(\{Y_i\}_{i\in I}\) is a collection of connected subspaces of \(X\) and \(\bigcap_{i\in I}Y_i\ne\emptyset\text{,}\) then \(Y=\bigcup_{i\in I}Y_i\) is connected.
If \(Y\subseteq X\) is connected, then any \(Y'\) satisfying \(Y\subseteq Y'\subseteq \overline{Y}\) is connected.
Proof.
Given a separation \(X=U_1\cup U_2\text{,}\) we have the separation \(Y=U_1'\cup U_2'\text{,}\) where \(U_i'=U_i\cap Y\text{.}\) Since \(Y\) is connected, we must have \(Y\subseteq U_1'\) or \(Y\subseteq U_2'\text{.}\) It follows that \(Y\subseteq U_1\) or \(Y\subseteq U_2\text{.}\)
Let \(Y=U_1\cup U_2\) be a separation. Take any \(y\in \bigcap_{i\in I}Y_i\text{.}\) We may assume without loss of generality that \(y\in U_1\text{.}\) Now since \(Y_i\) is connected for all \(i\) we must have \(Y_i\subseteq U_1\) or \(Y\subseteq U_2\) by (1). Since \(y\in Y_i\) and \(y\in U_1\text{,}\) it follows that \(Y_i\subseteq U_1\) for all \(i\in I\text{.}\) Thus \(Y=\bigcup_{i\in I}Y_i\subseteq U_1\text{.}\) We conclude that \(U_1=Y\text{,}\) showing that \(Y\) has no nontrivial separations. Thus \(Y\) is connected.
Assume \(Y\subseteq Y'\subseteq \overline{Y}\text{,}\) where \(Y\) is connected. Let \(Y'=C\cup D\text{,}\) where \(C\) and \(D\) are disjoint closed sets in \(Y'\text{.}\) By (1) we have \(Y\subseteq C\) or \(Y\subseteq D\text{.}\) Assume without loss of generality that \(Y\subseteq C\text{.}\) Since \(C\) is closed, the closure of \(Y\) in \(Y'\) is contained in \(C\text{.}\) But as we have seen in an earlier homework, the closure of \(Y\) in \(Y'\) is \(Y'\cap \overline{Y}=Y'\text{.}\) Thus \(Y'\subseteq C\text{,}\) or equivalently, \(C=Y'\text{.}\) We conclude that \(Y'\) is connected.
Theorem1.12.7.Connectedness and continuity.
If \(f\colon X\rightarrow Y\) is continuous and \(X\) is connected, then \(f(X)\) is connected.
Proof.
We identify \(f\) with the map \(f\colon X\rightarrow f(X)\) obtained by restricting the codomain, which is continuous and surjective. If \(U_1, U_2\) are disjoint open sets satisfying \(f(X)=U_1\cup U_2\text{,}\) then we have \(X=f^{-1}(f(X))=f^{-1}(U_1)\cup f^{-1}(U_2)\text{.}\) Since the sets \(f^{-1}(U_i)\) are open and disjoint, we have \(X=f^{-1}(U_1)\) or \(X=f^{-1}(U_2)\text{.}\) Since \(f\) is surjective, we have \(f(f^{-1}(U_i))=U_i\) for \(1\leq i\leq 2\text{.}\) It follows that \(U_1=f(X)\) or \(U_2=f(X)\text{.}\) We conclude \(f(X)\) is connected.
Corollary1.12.8.Graphs of continuous functions.
Let \(f\colon X\rightarrow Y\) be continuous and let \(\Gamma_f=\{(x,f(x))\in X\times Y\colon x\in X\}\text{,}\) the graph of \(f\text{.}\) If \(X\) is connected, then \(\Gamma_f\) is connected.
Proof.
Define \(F\colon X \rightarrow X\times Y\) as \(F(x)=(\id_X(x),f(x))=(x,f(x))\text{.}\) Since the two component functions \(\id_X\colon X\rightarrow X\) and \(f\colon X\rightarrow Y\) are continuous, so is \(F\text{.}\) From Theorem 1.12.7, we conclude that \(F(X)=\Gamma_f\) is continuous.
Theorem1.12.9.
Let \(\{X_i\}_{i\in I}\) be a collection of topological spaces, where \(I\) is nonempty, and let \(X=\prod_{i\in I}X_i\) be equipped with the product topology.
If \(X\) is connected, then \(X_i\) is connected for all \(i\in I\text{.}\)
If \(I\) is finite or countable then \(X\) is connected if and only if \(X_i\) is connected for all \(i\in I\text{.}\)
Assuming the axiom of choice, \(X\) is connected if and only if \(X_i\) is connected for all \(i\in I\text{.}\)
Proof.
We prove only (1) and (2). Proofs of (3) can be found in the literature: transfinite induction is often used.
Since each the projection map \(\pi_i\) is continuous for all \(i\in I\text{,}\) and since \(X_i=\pi_i(X_i)\text{,}\) we see using Theorem 1.12.7 that if \(X\) is connected, then \(X_i\) is connected.
Finite products.
Using induction it suffices to show that if \(X, Y\) are connected, then \(X\times Y\) is connected. First, choose any \(x_0\in X\) and note that the subspace \(\{x_0\}\times Y\text{,}\) being homeomorphic to \(Y\text{,}\) is connected. Similarly, for any \(y_0\in Y\text{,}\) the subspace \(X\times \{y_0\}\) is connected. It follows from (2) of Theorem 1.12.6 that the set \(T_{x_0,y_0}=\{x_0\}\times Y\cup X\times \{y_0\}\) is connected for any \((x_0,y_0)\in X\times Y\text{.}\) Lastly, since \(T_{x,y_0}\cap T_{x',y_0}\supseteq X\times\{y_0\}\ne \emptyset\) for any \(x,x'\in X\text{,}\) the union \(\bigcup_{x\in X}T_{x,y_0}\) is connected. But this union is \(X\) itself.
Countable product.
Let \(X=\prod_{n=1}^\infty X_n\) where \(X_n\) is connected for all \(n\text{.}\) Fix any element \(x=(x_1,x_2,\dots, )\in X\text{.}\) For each \(N\) the set
being homeomorphic to a finite product of connected spaces, is connected. Since \(X_N\cap X_{M}\ne \emptyset\) for all \(N, M\text{,}\) we conclude that \(Y=\bigcup_{N=1}^\infty X_N\) is connected.
Next, I claim that \(X=\overline{Y}\text{.}\) Indeed, given any \(z=(z_1,z_2,\dots, )\) and any open base element \(U=\prod_{n=1}^NU_n\times \prod_{n=N+1}^\infty X_n\) containing \(z\text{,}\) we have \((z_1,z_2,\dots, z_N, x_{N+1},x_{N+2},\dots)\in U\cap X_N\subseteq Y\text{.}\) Since connectedness is preserved under closure, we conclude that \(X\) is connected.
