Let \(\ts{X_\alpha}_{\alpha \in J}\) be an indexed family of connected spaces; let \(X\) be the product space
\[ X = \prod_{\alpha \in J} X_\alpha. \]
Let \(\mathbf{a} = (a_\alpha)\) be a fixed point of \(X\).
Given any finite subset \(K\) of \(J\), let \(X_K\) denote the subspace of \(X\) consisting of all points \(\mathbf{x} = (x_\alpha)\) such that \(x_\alpha = a_\alpha\) for \(\alpha \notin K\). Show that \(X_K\) is connected.
Show that the union \(Y\) of the spaces \(X_K\) is connected.
Show that \(X\) equals the closure of \(Y\); conclude that \(X\) is connected.
Solution
For finite \(K\subseteq J\), the space \(X_K\) is naturally homeomorphic to \[ \prod_{\alpha\in K}X_\alpha \] by forgetting the fixed coordinates. A finite product of connected spaces is connected, so \(X_K\) is connected.
Every \(X_K\) contains the fixed point \(\mathbf a\). Hence the union \[ Y=\bigcup_{K\subseteq J,\ K\text{ finite}}X_K \] is a union of connected sets with a common point, and is therefore connected.
Let \(U\) be a nonempty basic open set of the product \(X\). It restricts only finitely many coordinates, say those in \(K\). Choose \(x\in U\) and define \(y\) by \[ y_\alpha=x_\alpha\quad(\alpha\in K),\qquad y_\alpha=a_\alpha\quad(\alpha\notin K). \] Then \(y\in U\cap X_K\subseteq U\cap Y\). Thus every nonempty basic open set meets \(Y\), so \(\overline Y=X\). The closure of a connected set is connected, hence \(X\) is connected.