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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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"big. Really big. You just won't believe how vastly, hugely, mindbogglingly big it is." – Gerry Myerson Mar 6 '13 at 6:00
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
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
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 $\mathbb{Q}$. – arjafi 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
In general, any subspace of a separable metric space is also separable. – Haskell Curry Mar 6 '13 at 6:08

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