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Prove: the countable product of regular topological spaces is regular.

Label the countable product of $X_i$ as $X$. Given $x \in X$ and $U$ a closed set s.t. $ x \notin U$, let's find disjoint neighborhoods of $x$ and $U$. Because $U$ is closed in $X$, it's closed in each $X_i$ (label these closed sets as $U_i$. Also, each coordinate of $x$ is disjoint in each $U_i$, and from $X_i$'s regularity we get that there are open disjoint neighborhoods around $x$ and $U_i$ in $X_i$.

If we take the product of these neighborhoods we get what we want.

Is this proof correct? I'm new to the idea of product spaces so I'm not quite sure what I'm doing.

Martin Sleziak
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John H.
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2 Answers2

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There are multiple mistakes in your proof. First of all, the set $U$ is a subset of $X$, not a a subset of $X_i$. This mistake also cannot be fixed by considering the projection of $\pi_i(U)$ onto $X_i$, as $\pi_i(x)$ may be inside of $\pi_i(U)$. For example, consider a circle in $\mathbb{R}^2$ and the origin.

Also, I assume you want to consider the product topology on the product of spaces. However, the infinite product of open sets is in general not open in the product topology (see also Box Topology).

If you want a hint as to how to prove the assertion, take a look at this answer on MO.

Community
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Dominik
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Your proof does not work. The main reason is: it is possible that $x \notin U$, while, for all $i$, $x_i\in U_i$.

Here is one of the simplest ways to prove that the product of regular topological spaces is regular. Note that it works for the general case of product space (finite, countable or uncountable).

  1. Note that a space $X$ is regular if and only if for all $x\in X$ and all open sets $A$ in $X$, such that $x\in A$, there is an open set $B$ and a closed set $C$ in $X$, such that $x\in B\subseteq C \subseteq A$. (The proof is straight forward)

  2. Let $X$ be the product of $X_\lambda$, where, for each $\lambda$, $X_\lambda$ is regular. Let us prove $X$ is regular. Let $x\in X$ and $A$ be an open set in $X$ (in the product topology), such that $x \in A$. Then there is a basic open set $D$, such that $$x\in D \subseteq A$$ Since $D$ is a basic open set, we know that $D=\prod_\lambda D_\lambda$ where for all $\lambda$, $D_\lambda$ is open and, except for a FINITE number of indexes, $D_\lambda=X_\lambda$. Let $$\{\lambda_1,\cdots,\lambda_n\}=\{\lambda \,|\, D_\lambda \neq X_\lambda \}$$ Since for each $\lambda_k \in \{\lambda_1,\cdots,\lambda_n\}$, $X_{\lambda_k}$ is regular, and $x_{\lambda_k} \in D_{\lambda_k}$, there are open set $B_{\lambda_k}$ and a closed set $C_{\lambda_k}$ in $X_{\lambda_k}$, such that $x_{\lambda_k}\in B_{\lambda_k}\subseteq C_{\lambda_k} \subseteq D_{\lambda_k}$. Now, let $E$ be the product of open sets such that $E_\lambda=X_\lambda$, if $\lambda \notin \{\lambda_1,\cdots,\lambda_n\}$ and $E_{\lambda_k}=B_{\lambda_k}$, for $\lambda_k \in \{\lambda_1,\cdots,\lambda_n\}$. In a similar way, let $F$ be the product of closed sets such that $F_\lambda=X_\lambda$, if $\lambda \notin \{\lambda_1,\cdots,\lambda_n\}$ and $F_{\lambda_k}=C_{\lambda_k}$, for $\lambda_k \in \{\lambda_1,\cdots,\lambda_n\}$. It is a imediate that $E$ is an open set in $X$, $F$ is a closed set in $X$ and $$x\in E \subseteq F \subseteq D \subseteq A$$

Ramiro
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