# Product of Separable Spaces

In [Ross & Stone], it was shown that the product of $\leq$ continuum many separable spaces is separable while the product of more than continuum many separable spaces is not separable. My goal here is to write down a proof here as a record for future reference. While the product of continuum many separable factors is separable, it can never be hereditarily separable. Any product space with uncountably many factors, each of which has at least two points, has a subspace that is not separable. The non-separable subspace shown here is the $\Sigma-$product. The $\Sigma-$product of $\leq$ continuum (but uncountably) many separable spaces is an example of a space with the countable chain condition (ccc) that is not separable.

A. The Product of Continuum Many Separable Spaces
Suppose that $X_a$ is a separable space for each $a \in \mathbb{R}$. Then the product space $Y=\Pi_{a \in \mathbb{R}} X_a$ is separable.

Proof. The proof uses the fact that the real line $\mathbb{R}$ has a countable base. Let $\omega$ be the set of all natural numbers. For each $a \in \mathbb{R}$, let $\lbrace{x_{a,0},x_{a,1},x_{a,2},...}\rbrace$ be a countable dense subset of $X_a$. Let $\mathcal{B}$ be the collection of all open intervals in $\mathbb{R}$ with rational endpoints.

Let $F$ be a pairwise disjoint finite collection of $\mathcal{B}$. Fix a function $f:F \rightarrow \omega$. For this $(F,f)$ combination, we define a point $h \in Y$. For each open interval $B \in F$, $h(a)=x_{a,f(B)}$ for all points $a \in B$. For all other $a \in \mathbb{R}$, define $h(a)=x_{a,0}$. Since there are only countably many $(F,f)$ combinations, this algorithm defines a countable subspace $D$ of the product space $Y$. It follows that $D$ is dense in $Y$.

Comment
If there are $\mathcal{K}$ many factors where $\mathcal{K} \leq$ continuum, just use a subspace of the real line with cardinality $\mathcal{K}$ as an index set. Then choose a countable base of the index set and apply the same proof.

B. The Product of Greater Than Continuum Many Separable Spaces
Suppose that for each $a \in \mathbb{S}$, $X_a$ is a separable space with at least two points. If the product space $Y=\Pi_{a \in S} X_a$ is separable, then $\vert S \vert \leq \vert \mathbb{R} \vert$. Thus if the cardinality of the index set $S$ is greater than continuum, then the product space can never be separable.

Proof. For each $a \in S$, choose two disjoint nonempty open sets $U_{a,0},U_{a,1}$ in $X_a$. This is possible since $X_a$ has at least two points and is a Hausdorff space. Let $D$ be a countable dense subset of the product space $Y$. For each $a \in S$, define $D_a=\lbrace{h \in Y:h(a) \in U_{a,0}}\rbrace \cap D$. Each $D_a$ is nonempty because it is the intersection of an open set with a dense set. For $a \neq b$, $D_a \neq D_b$. So here is a one-to-one mapping of $S$ into the power set of the countable set $D$. Thus $\vert S \vert \leq \vert \mathbb{R} \vert$.

C. The Product Space With Uncountably Many Factors Is Not Hereditarily Separable

Proof. Suppose that we have a product space $Y=\Pi_{a \in S} X_a$ where $S$ is uncountable and each factor has at least two points. Given a point $p \in Y$, the $\Sigma-$product is the following subspace of the product space $Y$: $\Sigma_{a \in S} X_a=\lbrace{h \in Y:h(a) \neq p(a)}$ for only countably many $a \in S\rbrace$.

Let $\lbrace{h_0,h_1,h_3,...}\rbrace$ be a countable subset of $\Sigma_{a \in S} X_a$. Since each point $h_i$ disagrees with $p$ at only countably many points, choose one $b \in S$ such that each $h_i(b)=p(b)$. Then consider the open set $V=\lbrace{h \in \Sigma_{a \in S} X_a:h(b) \in W}\rbrace$ where $W$ is open in $X_b$ such that and $p(b) \notin W$. None of the $h_i$ belongs to $V$.

Comment
The $\Sigma-$product $\Sigma_{a \in S} X_a$ of $\leq$ continuum (but uncountably) many separable spaces is an example of “ccc and not separable”.

Reference
[Ross & Stone] Ross, K. A., Stone, A. H.  Product of Separable Spaces, The American Mathematical Monthly, Vol. 71, pp. 398-403.

## 2 thoughts on “Product of Separable Spaces”

1. ever4ys on said:

Thanks for the nice notes. Just want to point out that in part b, we need to assume X_a is Hausdorff, as it is apparent from the proof.