Product Spaces

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Definition and Properties of Product Spaces

Definition 1 Let a family of topological spaces \((X_i)_{i\in I}\) be given. Their product is the topological space on the product set \(X=\prod_{i\in I}X_i\) equipped with the initial topology with respect to the functions \(\pr_i:X\rightarrow X_i\).

Then by §Initial and Final Topology, ⁋Proposition 2, the product topology on \(X\) is the topology generated by the following collection as a subbase:

\[\mathcal{S}=\{\pr_i^{-1}(U_i)\mid U_i\text{ open in }X_i\}\]

Here, the base \(\mathcal{B}\) generated by \(\mathcal{S}\) consists of the sets

\[\prod_{i\in I} U_i,\qquad \text{$U_i$ open in $X_i$, $U_i=X_i$ for all but finitely many $i$}\]

For the same reason as in §Properties of Products, ⁋Proposition 3, taking products of spaces satisfies the associative law, and similarly there is a proposition for the commutative law. These are fairly obvious, so we do not state them separately.

On the other hand, applying §Initial and Final Topology, ⁋Proposition 3, we obtain the following.

Proposition 2 Let a product space \(X=\prod_{i\in I}X_i\) and a topological space \(Y\) be given, and suppose functions \(f_i:Y\rightarrow X_i\) are given. Then the function \(f=(f_i): Y\rightarrow X\) is continuous if and only if each \(f_i\) is continuous.

From this we obtain the following two corollaries.

Corollary 3 Let two product spaces \(X=\prod_{i\in I}X_i\), \(Y=\prod_{i\in I}Y_i\) sharing the same index set \(I\) be given, and suppose \(f_i:X_i\rightarrow Y_i\) are given. Then the map \(f:(x_i)\mapsto (f_i(x_i))\) is continuous if and only if each \(f_i\) is continuous.

For arbitrary sets \(X, Y\) and a function \(f:X \rightarrow Y\), the graph of \(f\), denoted \(\Gamma(f)\), is the subset of \(X\times Y\) given by

\[\Gamma(f)=\{(x,f(x)\mid x\in X\}\subseteq X\times Y\]

If both \(X, Y\) are topological spaces, then \(\Gamma(f)\) inherits the subspace topology from the product space \(X\times Y\).

Corollary 4 For topological spaces \(X,Y\) and a function \(f:X\rightarrow Y\), \(f\) is continuous if and only if the function \(g:x\mapsto (x,f(x))\) is a homeomorphism from \(X\) to \(\Gamma(f)\).

Moreover, we also know the obvious inverse of the above function \(g\),

\[\pr_X\vert_{\Gamma(f)}:\Gamma(f) \rightarrow X\]

In particular, since the constant map sending every point of \(X\) to \(y_0\in Y\) is continuous, by the above corollary we obtain the homeomorphism

\[X\rightarrow X\times\{y_0\}\]

Now consider an arbitrary subset \(A\) of \(X\times Y\). Then

\[A\cap (X\times \{y_0\})=\{(x,y)\mid (x,y)\in A,\quad y=y_0\}=\{(x,y_0)\mid (x,y_0)\in A\}\]

Since \(\Gamma(f)\) is equipped with the subspace topology of \(X\times Y\), the above set is open in \(\Gamma(f)\) if \(A\) is open in \(X\times Y\), and closed in \(\Gamma(f)\) if \(A\) is closed in \(X\times Y\). Therefore, applying Corollary 4 again and sending the above set to \(X\), the corresponding set is also open or closed. Let us denote this set by \(A(y_0)\subseteq X\). Of course, by a similar argument with the roles of \(X\) and \(Y\) interchanged, we can also obtain a subset \(A(x_0)\) of \(Y\).

Proposition 5 For any open set \(U\) in \(X\times Y\), \(\pr_X(U)\) and \(\pr_Y(U)\) are open in \(X\) and \(Y\), respectively.

Proof

This is obvious from the above argument and the following formulas:

\[\pr_X(U)=\bigcup_{y\in Y} U(y),\qquad \pr_Y(U)=\bigcup_{x\in X} U(x)\]

However, since an arbitrary union of closed sets need not be closed, the claim obtained by replacing \(A\) with a closed set in the above proposition does not hold. On the other hand, by Corollary 4, we also obtain the following proposition.

Proposition 6 If a function \(f:X_1\times X_2 \rightarrow Y\) is continuous at \((a_1,a_2)\in X_1\times X_2\), then the functions from \(X_1\) to \(Y\) and from \(X_2\) to \(Y\) defined by

\[x_1\mapsto f(x_1, a_2),\qquad x_2\mapsto f(a_1,x_2)\]

are continuous at \(x_1=a_1\) and \(x_2=a_2\), respectively.

However, the converse of this proposition does not hold.

Interior and Closure

Now we examine the relationship between product sets and interior and closure. First, taking closure behaves well with respect to product sets. That is, the following holds.

Proposition 7 Let a product space \(X=\prod_{i\in I} X_i\) and arbitrary subsets \(A_i\) of \(X_i\) be given. Then the following formula holds:

\[\prod_{i\in I} \cl A_i=\cl\left(\prod_{i\in I} A_i\right)\]

Proof

First, \(\prod_{i\in I}\cl A_i\) is a closed set. This follows from the formula

\[\prod_{i\in I}\cl A_i=\bigcap_{i\in I}\pr_i^{-1}(\cl A_i)\]

together with the fact that projections are continuous so each \(\pr_i^{-1}(\cl A_i)\) is closed, and that an arbitrary intersection of closed sets is closed. Then since \(\prod A_i\subseteq \prod\cl A_i\), by the minimality of closure we have

\[\cl\left(\prod_{i\in I}A_i\right)\subseteq \prod_{i\in I}\cl A_i\]

Conversely, suppose \(x=(x_i)\in\prod_{i\in I}\cl A_i\) is given, and consider an arbitrary neighborhood \(V\) of \(x\). Then, thinking of the base for the product topology, there exist open sets \(U_i\) such that \(x\in\prod U_i\subseteq V\) and \(U_i=X_i\) for all but finitely many \(i\). For each \(i\), since \(x_i\in \cl A_i\) and \(U_i\) is a neighborhood of \(x_i\), by §Interior, Closure, and Boundary, ⁋Proposition 6 we have \(U_i\cap A_i\neq\emptyset\), and we can choose an element \(a_i\in U_i\cap A_i\). Then \(a=(a_i)\) is an element of \(V\cap \prod A_i\), so every neighborhood of \(x\) meets \(\prod A_i\), and again by §Interior, Closure, and Boundary, ⁋Proposition 6 we have \(x\in \cl\left(\prod A_i\right)\).

However, the above proposition does not always hold for interior; it only holds when \(I\) is finite.

Proposition 8 Let a product space \(\prod_{i\in I} X_i\) with finite index set \(I\) and arbitrary subsets \(A_i\) of \(X_i\) be given. Then the following formula holds:

\[\prod_{i\in I} \interior A_i=\interior\left(\prod_{i\in I} A_i\right)\]

Proof

Since \(I\) is finite, \(\prod_{i\in I}\interior A_i\) is an open set belonging to the base for the product topology discussed above, and is contained in \(\prod A_i\). Therefore, by the maximality of interior,

\[\prod_{i\in I}\interior A_i\subseteq \interior\left(\prod_{i\in I}A_i\right)\]

Conversely, let \(x=(x_i)\in\interior\left(\prod A_i\right)\). Then, thinking of the base, there exist open sets \(U_i\) such that \(x\in \prod U_i\subseteq \prod A_i\). Fixing each \(j\in I\), for arbitrary \(y_j\in U_j\), the element obtained by replacing only the \(j\)-th component of \(x\) with \(y_j\) still belongs to \(\prod U_i\) (the other components satisfy \(x_i\in U_i\)), and hence to \(\prod A_i\). Reading the \(j\)-th component of this element, we have \(y_j\in A_j\). That is, \(U_j\subseteq A_j\), and since \(U_j\) is open, \(x_j\in U_j\subseteq \interior A_j\). Since this holds for every \(j\), we have \(x\in\prod_{i\in I}\interior A_i\).