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Let $A \subset R$, where $R$ is the real line. Let $F=\{f\mid f: A\rightarrow R^\omega \text{ is continuous}\}$. How big is the set $F$? Thanks for your help.

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6  
"big. Really big. You just won't believe how vastly, hugely, mindbogglingly big it is." –  Gerry Myerson Mar 6 '13 at 6:00
3  
Think of the national debt... –  copper.hat Mar 6 '13 at 6:00
    
@GerryMyerson Hitchhiker's guide rules! –  Henno Brandsma Mar 6 '13 at 15:54

1 Answer 1

up vote 8 down vote accepted

We are guaranteed of a countable dense subset $B \subset A$ since the reals are secound-countable (see the comments below).

Each $f$ is determined by its values at points in $B$, so $|F| \le |(\mathbb R^{\omega})^{B} | = \mathfrak c^{\aleph_0} = \mathfrak c$

But each constant function is continuous, so $|F| \ge \mathfrak c$. By Schroeder-Bernstein, $|F| = \mathfrak c$.

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Not sure the first part is that simple. What if $A=\Bbb R\setminus\Bbb Q$? –  Cameron Buie Mar 6 '13 at 6:04
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All subspaces of $\mathbb{R}$ are separable (follows from the second countability of $\mathbb{R}$), and so just use a countable dense subset of $F$ instead of $\mathbb{Q}$. –  Arthur Fischer Mar 6 '13 at 6:05
    
Ah! There we go. Although I suspect that Arthur meant a countable dense subset of $A$, rather than one of $F$. –  Cameron Buie Mar 6 '13 at 6:07
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In general, any subspace of a separable metric space is also separable. –  Haskell Curry Mar 6 '13 at 6:08

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