Why is $2^{\mathbb{R}}$ a separable function space?

Question

I know $\mathbb{N}^{\mathbb{R}}$ is a separable function space. Could someone please help me see why $2^{\mathbb{R}}$ is separable as well?

edit: I assume $2$ and $\mathbb{N}$ have the discrete topology and $2^{\mathbb{R}}$, $\mathbb{N}^{\mathbb{R}}$ have the product topology. And I was also wondering if $2^{\mathbb{R}}$ is compact and how to prove it.

@JoséCarlosSantos Sorry, should have mentioned: The discrete topology. — Math Learner, Jul 06 '17 at 17:29
A space with the discrete topology is compact if and only if it is finite, and it is separable if and only if it is countable. — Ben Grossmann, Jul 06 '17 at 17:48

score 2 · Answer 1 · answered Jul 06 '17 at 17:44

In my post I show (in the section powers of discrete spaces), that $D(\kappa)^{2^\kappa}$ has a dense set of size $\kappa$. (here $D(\kappa)$ is set of size $\kappa$ with the discrete topology). Now note that $|\mathbb{R}| = 2^{\aleph_0}$. So $\{0,1\}^{\mathbb{R}}$ is also separable, and it is compact, as a product of compact spaces (finite spaces are compact). This is a special case of Tychonoff's theorem.. For proofs, see any good text book.

All assuming you have the pointwise, i.e. product topology.

Why is $2^{\mathbb{R}}$ a separable function space?

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