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.
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