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I already figured out how to show that the countable product of separable topological spaces is separable, but I'm out of ideas when the index set has cardinality of $c$. My textbook says it is possible but gives no references. Any suggestions how to prove this statement?

In a less general setting, I would also be interested to see how a dense countable set is constructed to $\mathbb{R}^{\mathbb{R}}$. Thanks in advance.

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There is a related post on MO, in particular G. Edgar's comment seems to be useful. – Martin Sleziak Jan 20 '14 at 12:52
up vote 8 down vote accepted

This is a special case of Hewitt-Marczewski-Pondiczery theorem, see e.g. Theorem 2.3.15 in Engelking's General Topology:

If $d(X_s)\leq \alpha\geq\aleph_0$ for every $s\in S$ and $|S|\leq 2^\alpha$, then $d(\prod X_s)\leq\alpha$.

The $d(X)$ denotes the density of the topological space $X$, which is defined as $$d(X)=\min\{|D|; D\text{ is a dense subset of }X\}+\aleph_0.$$ I.e., $d(X)$ is the smallest cardinality of a dense subset, but if there is a finite dense subset, we put $d(X)=\aleph_0$.

This means that a topological space is separable if and only if $d(X)=\aleph_0$.

Some further references are given at Planetmath. Wikipedia article on separable space mentions Theorem 16.4c in Willard's General Topology as a reference for the special case you're asking about.

A proof of this theorem can be found at in this post from Ask a Topologist. (The post was written by Henno Brandsma.)

This theorem can be used to show that there is an independent family on $\mathbb N$ of cadinality $\mathfrak c$; see Stephan Geschke's MO post and paper.

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Thanks, I found from Willard's General Topology exactly the theorem that I was looking for. That a product of $c$ many separable spaces is separable. – T. Eskin Jan 8 '12 at 17:09
So the space $\{x\}$ is not separable? :-) – Asaf Karagila Jan 8 '12 at 17:46
@Asaf: I've added the correction that density is $\ge\aleph_0$ by definition. – Martin Sleziak Jan 8 '12 at 17:52

For $\mathbb{R}^\mathbb{R}$, try viewing it as the set of all functions from $\mathbb{R}$ to $\mathbb{R}$. Then the set of polynomials with rational coefficients is a countable dense subset. (Show that every nonempty open set contains such a polynomial. It might help to first show that every nonempty open set contains a polynomial with real coefficients.)

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Alright. Since the basic open sets in the product topology are of the form $\prod_{i\in \mathbb{R}} B_{i}$ where $B_{i}= \mathbb{R}$ except for finitely many $i_{1},...,i_{n}\in \mathbb{R}$, the only thing is then to show that there exists a rational polynomial that maps each $i_{k}$ to $B_{i_{k}}$?. For simplicity I guess I can assume that $i_{k}=k$ without any loss of generality? – T. Eskin Jan 8 '12 at 17:47
@ThomasE.: Right. Or to put it another way, given $x_1, \dots, x_n \in \mathbb{R}$ and open sets $B_1, \dots, B_n$, find a (rational) polynomial $p$ such that $p(x_i) \in B_i$ for each $i$. – Nate Eldredge Jan 8 '12 at 19:47
Alright, got it. Thanks. – T. Eskin Jan 9 '12 at 12:37
@NateEldredge Is $Q^R$ a dense subset in $R^R$? Where $Q$ is all the rationals in $R$. – Paul Aug 10 '12 at 7:08
@Paul: Yes. (Try proving it!) However, that doesn't help with this question as $\mathbb{Q}^\mathbb{R}$ is not countable. – Nate Eldredge Aug 10 '12 at 12:48

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