# Banach space and Hilbert space topology

Let $$B$$ be a Banach space. It is not necessarily true that there exists a Hilbert space $$H$$ linearly isometric to $$B$$.

However, is it true that there exists a Hilbert space $$H$$ homeomorphic to $$B$$?

• If $B$ is separable, then yes. All separable Banach Spaces are homeomorphic. So homeomorphic to $\ell^2$ Apr 7, 2019 at 21:35
• @user124910 We can extend this to non-separable as well. See my answer. Apr 7, 2019 at 21:37

Yes, but this is quite a deep result. Two infinite-dimensional Banach spaces $$X$$ and $$Y$$ are homeomorphic iff $$d(X)=d(Y)$$, where the density $$d(X)$$ is the minimal size of a dense subset of $$X$$.
So any separable infinite-dimensional Banach space is homeomorphic to the Hilbert space $$\ell^2$$ (and even to $$\mathbb{R}^\omega$$, because the result extends to locally convex completely metrisable TVS's as well). And for higher densities we have Hilbert spaces $$\ell_2(\kappa)$$ as models. Finite dimensional we only have the $$\mathbb{R}^n$$ up to homeomorphism, which are already Hilbert spaces.