Let \(I\) be a non empty index set, let \(\{X_\alpha\mid\alpha\in I\}\) be a family of topological spaces, and let \(A_\alpha\subset X_\alpha\) for each \(\alpha\).
Show that if \(A_\alpha\) is closed in \(X_\alpha\) for each \(\alpha\), then \(\prod A_\alpha\) is closed in \(\prod X_\alpha\).
Show that \(\overline{\prod A_\alpha}=\prod\overline{A_\alpha}\).
Prove or disprove: If \(A_\alpha\) is open in \(X_\alpha\) for each \(\alpha\), then \(\prod A_\alpha\) is open in \(\prod X_\alpha\).
Solution
Write \[ X=\prod_{\alpha\in I}X_\alpha, \qquad A=\prod_{\alpha\in I}A_\alpha, \] and let \(\pi_\alpha:X\to X_\alpha\) denote the coordinate projection.
1. If every \(A_\alpha\) is closed in \(X_\alpha\), then \(A\) is closed in \(X\).
For every \(\alpha\in I\), the coordinate projection \(\pi_\alpha\) is continuous in the product topology. Hence \[ \pi_\alpha^{-1}(A_\alpha) \] is closed in \(X\) whenever \(A_\alpha\) is closed in \(X_\alpha\). Moreover, \[ A =\prod_{\alpha\in I}A_\alpha =\bigcap_{\alpha\in I}\pi_\alpha^{-1}(A_\alpha). \] An arbitrary intersection of closed sets is closed, so \(A\) is closed in \(X\).
2. One always has\[ \overline{A}\subseteq\prod_{\alpha\in I}\overline{A_\alpha}. \]
For each \(\alpha\), \[ A_\alpha\subseteq\overline{A_\alpha}, \] so \[ A\subseteq\prod_{\alpha\in I}\overline{A_\alpha}. \] By 1, the product on the right is closed in \(X\) because every \(\overline{A_\alpha}\) is closed. Since \(\overline A\) is the smallest closed subset of \(X\) containing \(A\), the claimed inclusion follows.
3. Conversely,\[ \prod_{\alpha\in I}\overline{A_\alpha}\subseteq\overline A. \]
If the product on the left is empty, there is nothing to prove. Otherwise let \[ x=(x_\alpha)_{\alpha\in I}\in\prod_{\alpha\in I}\overline{A_\alpha}. \] Then every \(A_\alpha\) is nonempty. By the axiom of choice, the product of the nonempty sets \(A_\alpha\) is nonempty; choose a point \[ b=(b_\alpha)_{\alpha\in I}\in\prod_{\alpha\in I}A_\alpha. \]
Let \(U\) be any open neighborhood of \(x\) in \(X\). By the definition of the product topology, \(U\) contains a basic neighborhood \[ V=\bigcap_{\alpha\in F}\pi_\alpha^{-1}(U_\alpha), \] where \(F\subseteq I\) is finite and each \(U_\alpha\) is an open neighborhood of \(x_\alpha\) in \(X_\alpha\). Since \[ x_\alpha\in\overline{A_\alpha}, \] every \(U_\alpha\) with \(\alpha\in F\) meets \(A_\alpha\). Choose \[ a_\alpha\in U_\alpha\cap A_\alpha \qquad(\alpha\in F). \] Define \(y=(y_\alpha)_{\alpha\in I}\) by \[ y_\alpha= \begin{cases} a_\alpha,&\alpha\in F,\\b_\alpha,&\alpha\notin F.\end{cases} \] Then \(y\in A\) and \(y\in V\subseteq U\). Thus every neighborhood of \(x\) meets \(A\), so \(x\in\overline A\).
4. Therefore\[ \boxed{\overline{\prod_{\alpha\in I}A_\alpha} =\prod_{\alpha\in I}\overline{A_\alpha}}. \]
Combine 2 and 3.
5. The assertion in part (c) is false for an infinite index set.
Take \[ I=\NN, \qquad X_n=\RR, \qquad A_n=(-1,1) \] for every \(n\in\NN\). Each \(A_n\) is open in \(X_n\). Let \[ A=\prod_{n\in\NN}(-1,1)\subseteq\RR^{\NN} \] with the product topology. The zero sequence \(0=(0,0,\ldots)\) lies in \(A\).
Suppose \(A\) were open. Then some basic open neighborhood \(V\) of \(0\) would satisfy \[ 0\in V\subseteq A. \] A basic neighborhood in the product topology restricts only finitely many coordinates, so there is a finite set \(F\subset\NN\) such that all coordinates outside \(F\) are unrestricted. Choose \(m\notin F\) and define \(y\in\RR^{\NN}\) by \[ y_m=2, \qquad y_n=0\quad(n\ne m). \] Then \(y\in V\), because its only nonzero coordinate is unrestricted, but \(y\notin A\) because \(y_m=2\notin(-1,1)\). This contradicts \(V\subseteq A\). Hence \(A\) is not open.
For comparison, if \(I\) is finite, then a product of open subsets is open; more generally, a product \(\prod A_\alpha\) with all \(A_\alpha\) open is a basic open set whenever \(A_\alpha=X_\alpha\) for all but finitely many \(\alpha\).