# Sigma-products of first countable spaces

A product space is never first countable if there are uncountably many factors. For example, $\prod_{\alpha < \omega_1}\mathbb{R}=\mathbb{R}^{\omega_1}$ is not first countable. In fact any dense subspace of $\mathbb{R}^{\omega_1}$ is not first countable. In particular, the subspace of $\mathbb{R}^{\omega_1}$ consisting of points which have at most countably many non-zero coordinates is not first countable. This subspace is called the $\Sigma$-product of $\omega_1$ many copies of the real line $\mathbb{R}$ and is denoted by $\Sigma_{\alpha<\omega_1} \mathbb{R}$. However, this $\Sigma$-product is a Frechet space (or a Frechet-Urysohn space). In this post, we show that the $\Sigma$-product of first countable spaces is a Frechet space.

Consider the product space $X=\prod_{\alpha \in A} X_\alpha$. Fix a point $a \in X$. Consider the following subspace of $X$:

$\Sigma_{\alpha \in A} X_\alpha(a)=\left\{x \in X: x_\alpha \ne a_\alpha \text{ for at most countably many } \alpha \in A \right\}$

The above subspace of $X$ is called the $\Sigma$-product of the spaces $\left\{X_\alpha: \alpha \in A \right\}$ about the base point $a$. When the base point is understood, we simply say the $\Sigma$-product of the spaces $\left\{X_\alpha: \alpha \in A \right\}$ and use the notation $\Sigma_{\alpha \in A} X_\alpha$ to denote the space.

For each $y \in \Sigma_{\alpha \in A} X_\alpha$, define $S(y)$ to be the set of all $\alpha \in A$ such that $y_\alpha \ne a_\alpha$, i.e., the support of the point $y$. Another notion of support is that of standard basic open sets in the product topology. A standard basic open set is a set $O=\prod_{\alpha \in A} O_\alpha$ where each $O_\alpha$ is an open subset of $X_\alpha$. The support of $O$, denoted by $supp(O)$ is the finite set of all $\alpha \in A$ such that $O_\alpha \ne X_\alpha$.

A space $Y$ is said to be first countable if there exists a countable local base at each point in $Y$. A space $Y$ is said to be a Frechet space if for each $y \in Y$ and for each $M \subset Y$, if $y \in \overline{M}$, then there exists a sequence $\left\{y_n: n=1,2,3,\cdots \right\}$ of points of $M$ such that the sequence converges to $y$. Frechet spaces also go by the name of Frechet-Urysohn spaces. Clearly, any first countable space is Frechet. The converse is not true (see Example 1 in this post). We prove the following theorem.

Theorem 1

Suppose each factor $X_\alpha$ is a first countable space. Then the $\Sigma$-product $\Sigma_{\alpha \in A} X_\alpha$ is a Frechet space.

Proof of Theorem 1
Let $\Sigma=\Sigma_{\alpha \in A} X_\alpha$. Let $M \subset \Sigma$ and let $x \in \overline{M}$. We proceed to define a sequence of points $t_n \in M$ such that the sequence $t_n$ converges to $x$. For each $\alpha \in A$, choose a countable local base $\left\{B_{\alpha,j}: j=1,2,3,\cdots \right\}$ at the point $x_\alpha \in X_\alpha$. Assume that $B_{\alpha,1} \supset B_{\alpha,2} \supset B_{\alpha,3} \supset \cdots$. Then enumerate the countable set $S(x)$ by $S(x)=\left\{\beta_{1,1},\beta_{1,2},\beta_{1,3},\cdots \right\}$. Let $C_1=\left\{\beta_{1,1} \right\}$. The following set $O_1$ is an open subset of $\Sigma$.

$O_1=\biggl(\prod_{\alpha \in C_1} B_{\alpha,1} \times \prod_{\alpha \in A-C_1} X_\alpha \biggr) \cap \Sigma$

Note that $O_1$ is an open set containing $x$. Choose $t_2 \in O_1 \cap M$. Enumerate the support $S(t_2)$ by $S(t_2)=\left\{\beta_{2,1},\beta_{2,2},\beta_{2,3},\cdots \right\}$. Form the finite set $C_2$ by picking the first two points of $S(x)$ and the first two points of $S(t_2)$, i.e., $C_2=\left\{\beta_{1,1},\beta_{1,2},\beta_{2,1},\beta_{2,2} \right\}$. Then form the following open subset of $\Sigma$.

$O_2=\biggl(\prod_{\alpha \in C_2} B_{\alpha,2} \times \prod_{\alpha \in A-C_2} X_\alpha \biggr) \cap \Sigma$

Choose $t_3 \in O_2 \cap M$. Enumerate the support $S(t_3)$ by $S(t_3)=\left\{\beta_{3,1},\beta_{3,2},\beta_{3,3},\cdots \right\}$. Then let $C_3=\left\{\beta_{1,1},\beta_{1,2},\beta_{1,3},\ \beta_{2,1},\beta_{2,2},\beta_{2,3},\ \beta_{3,1},\beta_{3,2},\beta_{3,3} \right\}$, i.e., picking the first three points of $S(x)$, the first three points of $S(t_2)$ and the first three points of $S(t_3)$. Now, form the following open subset of $\Sigma$.

$O_3=\biggl(\prod_{\alpha \in C_3} B_{\alpha,3} \times \prod_{\alpha \in A-C_3} X_\alpha \biggr) \cap \Sigma$

Choose $t_4 \in O_2 \cap M$. Let this inductive process continue and we would obtain a sequence $t_2,t_3,t_4,\cdots$ of points of $M$. We claim that the sequence converges to $x$. Before we prove the claim, let’s make a few observations about the inductive process of defining $t_2,t_3,t_4,\cdots$. Let $C=\bigcup_{j=1}^\infty C_j$.

• Each $C_j$ is the support of the open set $O_j$.
• The sequence of open sets $O_j$ is decreasing, i.e., $O_1 \supset O_2 \supset O_3 \supset \cdots$. Thus for each integer $j$, we have $t_k \in O_j$ for all $k \ge j$.
• The support of the point $x$ is contained in $C$, i.e., $S(x) \subset C$.
• The support of the each $t_j$ is contained in $C$, i.e., $S(t_j) \subset C$.
• In fact, $C=S(x) \cup S(t_2) \cup S(t_3) \cup \cdots$.
• The previous three bullet points are clear since the inductive process is designed to use up all the points of these supports in defining the open sets $O_j$.
• Consequently, for each $j$, $x_\alpha=(t_j)_\alpha=a_\alpha$ for each $\alpha \in A-C$. In other words, $x$ and each $t_j$ agree (and agree with the base point $a$) on the coordinates outside of the countable set $C$.

Let $U=\prod_{\alpha \in A} U_\alpha$ be a standard open set in the product space $X=\prod_{\alpha \in A} X_\alpha$ such that $x \in U$. Let $U^*=U \cap \Sigma$. We show that for some $n$, $t_j \in U^*$ for all $j \ge n$.

Let $F=supp(U)$ be the support of $U$. Let $F_1=F \cap C$ and $F_2=F \cap (A-C)$. Consider the following open set:

$U^{**}=\biggl(\prod_{\alpha \in C} U_\alpha \times \prod_{\alpha \in A-C} X_\alpha \biggr) \cap \Sigma$

Note that $supp(U^{**})=F_1$. For each $\alpha \in F_1$, choose $B_{\alpha,k(\alpha)} \subset U_\alpha$. Let $m$ be the maximum of all $k(\alpha)$ where $\alpha \in F_1$. Then $B_{\alpha,m} \subset U_\alpha$ for each $\alpha \in F_1$. Choose a positive integer $p$ such that:

$F_1 \subset W=\left\{\beta_{i,j}: i \le p \text{ and } j \le p \right\}$

Let $n=\text{max}(m,p)$. It follows that there exists some $n$ such that $O_n \subset U^{**}$. Then $t_j \in U^{**}$ for all $j \ge n$. It is also the case that $t_j \in U^{*}$ for all $j \ge n$. This is because $x=t_j$ on the coordinates not in $C$. $\blacksquare$

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$\copyright \ 2014 \text{ by Dan Ma}$