Showing that two Banach spaces are homeomorphic when their dimensions are equal.

Let $X$ and $Y$ be Banach spaces. It is quite easy to show that they are homeomorphic when their dimensions are finite and equal.

However, I find it difficult to show that they are homeomorphic when their dimensions are infinite and equal. Here, the statement that their dimensions are equal must mean that a basis of $X$ and a basis of $Y$ have the same cardinality. (I know that any vector space has a basis by Zorn's Lemma.)

Could anyone help me how to show that they are homeomorphic when their dimensions are infinite?

Solution 1:

Let me comment on that question. The algebraic dimension for Banach spaces is actually rather useless. It seems that the correct dimension should be

$${\rm correct\, dimension} = \left\{ \begin{array}{ll}\dim X,& X\text{ is finite-dimensional},\\ \min |A|\colon A\subseteq X\text{ is dense},& \text{otherwise}.\end{array}\right.$$

Theorem. Let $X$ and $Y$ be Banach spaces with the same dimension, as defined above. Then $X$ and $Y$ are homeomorphic.

References:

1. M. I. Kadets, Proof of the topological equivalence of all separable infinite-dimensional Banach spaces, Functional Analysis and Its Applications, 1 (1967), 53–62.

2. H. Toruńczyk, Characterizing Hilbert space topology, Fund. Math. 111 (1981), 247–262.

Solution 2:

Both $\ell_2$ and $\ell_\infty$ have the same algebraic dimension of $2^{\aleph_0}$. But they are clearly not homeomorphic, since only one separable.