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Let $\{X_j\}_{j=1}^{+\infty}$ be a sequence of separable spaces.

The goal is to prove that the infinite cartesian product of separable spaces is indeed separable by showing that the product has a dense subset, arising from the fact that each $X_m$ has a countably dense subset.

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up vote 11 down vote accepted

For $n\in\Bbb Z^+$ let $D_n$ be a countable dense subset of $X_n$, and fix a point $x_n\in D_n$. For $m\in\Bbb Z^+$ let

$$E_m=\left\{y\in\prod_{n\in\Bbb Z^+}D_n:y_n=x_n\text{ for all }n\ge m\right\}\;,$$

and let $$E=\bigcup_{n\in\Bbb Z^+}E_m\;.$$

Then each $$E_m=\prod_{1\le n<m}D_n\times\prod_{n\ge m}\{x_n\}$$ is clearly countable, so $E$ is countable. Every non-empty open set in $X=\prod_{n\in\Bbb Z^+}X_n$ contains a basic open set of the form

$$B=\prod_{1\le n<m}V_n\times\prod_{n\ge m}X_n\;,$$ where $V_n$ is a non-empty open set in $X_n$ for $1\le n<m$, and clearly $B\cap E_m\ne\varnothing$, so $E$ is dense in $X$.

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How to know that every non-empty open set in $X$ contains a basic open set of the form stated above? – Idonknow Dec 6 '14 at 8:59
@Idonknow: That’s the definition of the product topology. – Brian M. Scott Dec 6 '14 at 20:15

Let for each $k$, $D_j$ countable and dense in $X_j$ $$D:=\bigcup_{n\geqslant 1}\{(x_1,\dots,x_n,x_{n+1},\dots),x_j\in D_j, 1\leqslant j\leqslant n\}.$$ Then:

  • $D$ is countable as an union of finite cartesian products of countable sets.
  • $D$ is dense in $X$. If $x\in X$ and $V$ is a neighborhood of $x$, by definition of the product topology, we can find an integer $N$ and for $1\leqslant j\leqslant N$, an open set $O_j$ of $X_j$ with $x_j\in O_j$. Now we use density of $D_j$ in $X_j$.
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