I was trying to prove the claim that separable infinite dimensional Banach spaces are all homeomorphic, but I fell into a trap that relies on the Continuum Hypothesis.
My reasoning was as follows:
Call the space $X$. By Baire $X$ has dimension $D > \aleph_0$.
The number of points we can get to via sequences of the countable dense subset is at most $\aleph_0^{\aleph_0}$ = $|\mathbb{R}|$.
There are $|\mathbb{R}|$ points in each dimension of $X$.
Each point is a countable sum of points on at most $\aleph_0$ dimensions.
So there are at most $|\mathbb{R}| D^{\aleph_0}$ points in $X$.
So $|\mathbb{R}| \geq |\mathbb{R}| D^{\aleph_0} = |\mathbb{R}| D$ points in $X$.
So we have $D \leq |\mathbb{R}|$.
But now I don't know if there's a way to fix this argument to show that $D = |\mathbb{R}|$ without assuming CH. I also would like to avoid using the high powered theorems typically needed for this proof.
Best Answer
See Cardinality of a Hamel basis of $\ell_1(\mathbb{R})$. The answer by t.b. cites a half-page paper that gives an elementary proof of the desired result: