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}

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