Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
There is a related post on MO, in particular G. Edgar's comment seems to be useful. –  Martin Sleziak Jan 20 at 12:52
    

2 Answers 2

up vote 7 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.

share|improve this answer
    
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. –  Thomas E. 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.)

share|improve this answer
    
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? –  Thomas E. 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. –  Thomas E. 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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.