why the space $I^{(2^C)}$ is not separable

Question

I have the space $I^{(2^{\mathfrak{c}})}$ where $I= [0,1]$ and $\mathfrak{c}$ is $| \mathbb R |$.

My professor wrote in the lecture that the space is not separable and I would like to know why not.

Can you explain it to me?

Answer 1

Cardinal functions $d$ and $w$ can help here:

Let $d(X)$ (the density of $X$) be the smallest infinite cardinality of a dense subset of $X$. (well-defined as cardinal numbers are well-ordered and $|X|$ is an upperbound.)

So $X$ separable iff $d(X) = \aleph_0$.

Let $w(X)$ (the weight of $X$) be the smallest infinite cardinality of an open base for $X$. (again well-defined as $|\mathcal{T}|$ is an upper bound.)

Lemma 1: In a regular space $X$ we have $w(X) \le 2^{d(X)}$.

Proof: fix a dense subset $D$ of size $d(X)$ and let $\mathcal{B}$ be any open base of $X$. Then the map $f: \mathcal{B} \to \mathscr{P}(D)$ defined by $f(O) = O \cap D \,(\subseteq D)$ is an injection for regular spaces. Hence $$w(X) \le |\mathcal{B}| \le |\mathscr{P}(D)| = 2^{d(X)}$$

Lemma 2: for $X$ a $T_0$ space: $|X| \le 2^{w(X)}$.

Proof: let $\mathcal{B}$ be an open base for $X$. Then the map $f: X \to \mathscr{P}(\mathcal{B})$ defined by $f(x) = \{O \in \mathcal{B}: x \in O\}$ is 1-1 (for $T_0$ spaces) and so $|X| \le |\mathscr{P}(\mathcal{B})| = 2^{w(X)}$.

A consequence of the previous two lemmas for the interesting case $d(X) = \aleph_0$:

A separable $T_3$ space $X$ can have size at most $2^{2^{\aleph_0}} =2^{\mathfrak{c}}$.

As $$|I^{2^\mathfrak{c}}| = \mathfrak{c}^{2^\mathfrak{c}} \ge 2^{2^{\mathfrak{c}}} > 2^{\mathfrak{c}}$$ (last step by Cantor's theorem; $\ge$ can really be $=$) this means your space cannot be separable: it's simply to big for it.