The product of first countable spaces

All spaces under consideration are Hausdorff. First countable spaces are those spaces where there is a countable local base at every point in the space. This is quite a strong property. For example, every first countable space that is also compact has a cap on its cardinality and the cap is the cardinality of the real line (the continuum). See The cardinality of compact first countable spaces, I in this blog. In fact, if the compact and first countable space is uncountable, it has cardinality continuum (see The cardinality of compact first countable spaces, III). Any metric space (or metrizable space) is first countable. In this post, we discuss the product of first countable spaces. In this regard, first countable spaces and metrizable spaces behave similarly. We show that the product of countably many first countable spaces is first countable while the product of uncountably many first countable is not first countable. For more information on the product topology, see [2].

The Product Space
Consider a collection of sets $A_\alpha$ where $\alpha \in S$. Let $W=\bigcup \limits_{\alpha \in S} A_\alpha$. The product $\prod \limits_{\alpha \in S} A_\alpha$ is the set of all functions $f:S \mapsto W$ such that for each $\alpha \in S$, $f(\alpha) \in A_\alpha$. If the index set $S=\left\{1,2,\cdots,n\right\}$ is finite, the functions $f$ can be regarded as sequences $(f_1,f_2,\cdots,f_n)$ where each $f_i \in A_i$. If the index set $S=\mathbb{N}$, we can think of elements $f$ of the product as the sequence $(f_1,f_2,\cdots)$ where each $f_i \in A_i$. In general we can regard $f \in \prod \limits_{\alpha \in S} A_\alpha$ as functions $f:S \mapsto W$ or as sequences $f=(f_\alpha)_{\alpha \in S}$.

Consider the topological spaces $X_\alpha$ where $\alpha \in S$. Let $X=\prod \limits_{\alpha \in S} X_\alpha$ be the product as defined above. The product space of the spaces $X_\alpha$ is $X$ with the topology defined in the following paragraph.

Let $\tau_\alpha$ be the topology of each space $X_\alpha$, $\alpha \in S$. Consider $Y=\prod \limits_{\alpha \in S} O_\alpha$ where for each $\alpha \in S$, $O_\alpha \in \tau_\alpha$ (i.e. $O_\alpha$ is open in $X_\alpha$) and $O_\alpha=X_\alpha$ for all but finitely many $\alpha \in S$. The set of all such sets $Y$ is a base for a topology on the product $X=\prod \limits_{\alpha \in S} X_\alpha$. This topology is called the product topology of the spaces $X_\alpha$, $\alpha \in S$.

To more effectively work with product spaces, we consider a couple of equivalent bases that we can define for the product topology. Let $\mathcal{B}_\alpha$ be a base for the space $X_\alpha$. Consider $B=\prod \limits_{\alpha \in S} B_\alpha$ such that there is a finite set $F \subset S$ where $B_\alpha \in \mathcal{B}_\alpha$ for each $\alpha \in F$ and $B_\alpha=X_\alpha$ for all $\alpha \in S-F$. The set of all such sets $B$ is an equivalent base for the product topology.

Another equivalent base is defined using the projection maps. For each $\alpha \in S$, consider the map $\pi_\alpha:\prod \limits_{\beta \in S} X_\beta \mapsto X_\alpha$ such that $\pi_\alpha(f)=f_\alpha$ for each $f$ in the product. In words, the function $\pi_\alpha$ maps each point in the product space to its $\alpha^{th}$ coordinate. The mapping $\pi_\alpha$ is called the $\alpha^{th}$ projection map. For each set $U \subset X_\alpha$, $\pi_\alpha^{-1}(U)$ is the following set:

$\pi_\alpha^{-1}(U)=\left\{f \in \prod \limits_{\beta \in S} X_\beta: \pi_\alpha(f)=f_\alpha \in U\right\}$

Consider sets of the form $\bigcap \limits_{\alpha \in F} \pi_\alpha^{-1}(U_\alpha)$ where $F \subset S$ is finite and $U_\alpha$ is open in $X_\alpha$ for each $\alpha \in F$. The set of all such sets is another equivalent base for the product topology. If we only require that each $U_\alpha \in \mathcal{B}_\alpha$, a predetermined base for the coordinate space $X_\alpha$, we also obtain an equivalent base for the product topology.

Countable Product of First Countable Spaces
For $i=1,2,3,\cdots$, let $X_i$ be a first countable space. We show that $X=\prod \limits_{i=1}^{\infty}X_i$ is a first countable space.

Let $\mathbb{N}$ be the set of positive integers. For each $n \in \mathbb{N}$, let $[n]=\left\{1,2,\cdots,n\right\}$ and let $\mathbb{N}^{[n]}$ be the set of all functions $t:[n] \mapsto \mathbb{N}$. For each $i$ and each $x \in X_i$, let $\mathcal{B}_x(i)=\left\{B_x(i,j): j \in \mathbb{N}\right\}$ be a countable local base at $x$.

Let $f \in X=\prod \limits_{i=1}^{\infty}X_i$. We wish to define a countable local base at $f$. For each $n \in \mathbb{N}$, define $W_n$ to be:

$W_n=\left\{\prod \limits_{i=1}^n B_{f(i)}(i,t(i)):t \in \mathbb{N}^{[n]}\right\}$

Let $W$ be the set of all subsets of the product space $X$ of the following form:

$\prod \limits_{j=1}^{\infty}V_j$ where there is some $n \in \mathbb{N}$ such that $\prod \limits_{j=1}^{n}V_j \in W_n$ and for all $j>n$, $V_j=X_j$.

Each $W_n$ is countable and $W$ is essentially the union of all the $W_n$. Thus $W$ is countable. We claim that $W$ is a local base at $f$. Let $O \subset X$ be an open set containing $f$. We can assume that $O=\prod \limits_{i=1}^\infty O_i$ where there is some $n \in \mathbb{N}$ such that for each $i \le n$, $O_i$ is open in $X_i$ and for $i>n$, $O_i=X_i$.

For each $i \le n$, $f(i) \in O_i$. Choose some $B_{f(i)}(i,t(i))$ such that $f(i) \in B_{f(i)}(i,t(i)) \subset O_i$. Let $V=\prod \limits_{i=1}^{\infty} V_i$ such that $\prod \limits_{i=1}^{n} V_i=\prod \limits_{i=1}^n B_{f(i)}(i,t(i))$ and $V_i=X_i$ for all $i>n$. Then $V \in W$ and $f \in V \subset O$. This completes the proof that $X=\prod \limits_{i=1}^{\infty}X_i$ is a first countable space.

Uncountable Product
Let $S$ be an uncountable index set. For $\alpha \in S$, let $X_\alpha$. We want to avoid the situation that all but countably many $X_\alpha$ are one-point space. So we assume each coordinate space $X_\alpha$ has at least two points, say, $p_\alpha$ and $q_\alpha$ with $p_\alpha \ne q_\alpha$. We show that $X=\prod \limits_{\alpha \in S}X_\alpha$ is not first countable.

Let $f \in \prod \limits_{\alpha \in S}X_\alpha$. Let $U_1,U_2, \cdots$ be open subsets of the product space such that for each $i$, $f \in U_i$. We show that there is some open set $O$ such that $f \in O$ and each $U_i \nsubseteq O$. For each $i$, there is a basic open set $B_i=\bigcap \limits_{\alpha \in F_i} \pi_\alpha^{-1}(U_{\alpha,i})$ such that $f \in B_i \subset U_i$.

Let $F=F_1 \cup F_2 \cup \cdots$. Since $S$ is uncountable and $F$ is countable, choose $\gamma \in S-F$. Since $X_\gamma$ has at least two points $p_\gamma$ and $q_\gamma$, choose one of them that is different from $f_\gamma$, say, $p_\gamma$. Choose two disjoint open subsets $M_1$ and $M_2$ of $X_\gamma$ such that $f_\gamma \in M_1$ and $p_\gamma \in M_2$ of $X_\gamma$. Let $O=\prod \limits_{\alpha \in S}O_\alpha$ such that $O_\gamma=M_1$ and $O_\alpha=X_\alpha$ for all $\alpha \ne \gamma$. We have $f \in O$. For each $i$, there is $g_i \in B_i \subset U_i$ such that $g_i(\gamma)=p_\gamma$. Thus each $g_i \notin O$. Thus there is no countable local base at $f$. Thus any product space with uncountably many factors, each of which has at least two points, is never first countable.

Reference

1. Engelking, R. General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

1 thought on “The product of first countable spaces”

1. “Choose two disjoint open subsets M_1 and M_2 of X_\gamma such that f_\gamma \in M_1 and p_\gamma \in M_2 of X_\gamma” – It is possible that two disjoin open subsets do not exist in X_\gamma.