Define the term “connected component” for a topological space. Prove that a connected component is connected.
Solution
Fix \(x\in X\) and let \[ \mathcal C_x =\{C\subseteq X:C\text{ is connected and }x\in C\}. \] Define the connected component of \(x\) to be \[ K_x=\bigcup_{C\in\mathcal C_x}C. \]
1. The family \(\mathcal C_x\) is nonempty.
The singleton \(\{x\}\) is connected and contains \(x\), so \[ \{x\}\in\mathcal C_x. \]
2. The set \(K_x\) is connected.
Suppose, for contradiction, that \[ K_x=U\cup V \] is a separation of \(K_x\), so \(U\) and \(V\) are disjoint nonempty sets open in the subspace \(K_x\). By 1, \(x\in K_x\). Since \(U\) and \(V\) are disjoint and cover \(K_x\), exactly one contains \(x\); after interchanging them if necessary, assume \[ x\in U. \]
Take any \(C\in\mathcal C_x\). Then \[ C=(C\cap U)\cup(C\cap V), \] where \(C\cap U\) and \(C\cap V\) are disjoint sets open in the subspace \(C\). Moreover, \[ x\in C\cap U, \] so \(C\cap U\) is nonempty. Because \(C\) is connected, \(C\cap V\) must therefore be empty. Hence \[ C\subseteq U \] for every \(C\in\mathcal C_x\). Taking the union over all such \(C\) gives \[ K_x\subseteq U, \] so \(V=\varnothing\), contradicting that \(U,V\) form a separation. Thus \(K_x\) is connected.
3. The set \(K_x\) is maximal among connected subsets of \(X\) containing \(x\).
Let \(D\subseteq X\) be connected with \[ K_x\subseteq D. \] Since \(x\in K_x\), we have \(x\in D\), and therefore \[ D\in\mathcal C_x. \] By the definition of \(K_x\) as the union of all members of \(\mathcal C_x\), \[ D\subseteq K_x. \] Together with \(K_x\subseteq D\), this gives \(D=K_x\). Thus no strictly larger connected subset containing \(x\) exists.
4. Consequently, connected components are exactly the maximal connected subsets of \(X\), and every connected component is connected.
By <1>2–<1>3, \(K_x\) is connected and maximal connected for every \(x\in X\). Conversely, if \(M\subseteq X\) is maximal connected and \(x\in M\), then \(M\in\mathcal C_x\), so \[ M\subseteq K_x. \] Since \(K_x\) is connected by 2 and contains \(M\), maximality of \(M\) gives \[ M=K_x. \] Hence the two standard descriptions agree, and in particular every connected component is connected.