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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?

Tomasz Kania
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Keith
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  • Are you sure it is valid for infinite dimensional spaces. Take $\mathcal{C}^0([0,1])$ equipped with $|\cdot|\infty$ and $\mathcal{C}^1([0,1])$ with $|f|{\mathcal{C}^1}=\int_0^1(|f(t)|+|f'(t)|)dt$ – marwalix Jun 05 '15 at 05:05
  • @marwalix, all separable infinite-dimensional Banach spaces are homeomorphic although the homeomorphism is usually not linear. – Tomasz Kania Jun 05 '15 at 13:36
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    They are homeomorphic iff they have the same topological density (e.g. both separable). But the proof is quite long. It's relatively easy for the standard $\ell_p$ spaces etc. – Henno Brandsma Jun 06 '15 at 07:25

2 Answers2

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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.

Tomasz Kania
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    (1) Stupid seems a bit excessive. (2) Why not define it as the least $|A|$ for which $\operatorname{span}(A)$ is dense in $X$? You get both the vector space connection, and the finite dimensional case as well. Otherwise dimension is not even related to the vector space structure. Just the metric structure. – Asaf Karagila Jun 05 '15 at 14:15
  • Asaf, they are the same. – Tomasz Kania Jun 05 '15 at 14:17
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    I know they are the same. I meant that the definition I suggested don't treat Banach spaces just as metric spaces, but as actual vector spaces. (And I stand by the first part of the comment that "stupid" is somewhat excessive here.) – Asaf Karagila Jun 05 '15 at 14:18
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    I believe that for non-separable (and so automatically infinite-dimensional, in whatever sense) Banach spaces $X,Y$ we also have that $d(X) = d(Y)$ iff $X$ is homeomorphic to $Y$, where $d(X)$ is (as usual) the minimal cardinality of a dense subset. I think it's in Bessaga and Peczynski. It's quite non-trivial :) – Henno Brandsma Jun 06 '15 at 07:23
  • @Henno Brandsma, thank you, you are completely right. Corrected. – Tomasz Kania Jun 06 '15 at 08:44
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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.

Asaf Karagila
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