Is a Hilbert space determined by its algebraic dimension? [duplicate]

Solution 1:

No, Hilbert spaces of different Hilbert dimensions can have the same algebraic dimension as vector spaces over $\mathbb R$ (or $\mathbb C$, take your pick).

For a cardinal $\kappa$, let $H_{\kappa}$ be the $\kappa$-dimensional real (or complex, whatever) Hilbert space, i.e., it has an orthonormal basis of cardinality $\kappa$. Let $\mathcal H=\{H_{\kappa}:\aleph_0\le\kappa\le2^{\aleph_0}\}$. The set $\mathcal H$ contains at least two nonisomorphic Hilbert spaces ($H_{\aleph_0}$ and $H_{2^{\aleph_0}}$) and maybe as many as $2^{\aleph_0}$ of them depending on your set theory. I claim that they are all algebraically isomorphic because they all have algebraic dimension $2^{\aleph_0}$ as vector spaces over $\mathbb R$.

First, since the number of points in the space $H_{\kappa}\in\mathcal H$ with orthonormal basis $\mathcal B$ is $|H_{\kappa}|\le|\mathcal B|^{\aleph_0}2^{\aleph_0}\le(2^{\aleph_0})^{\aleph_0}2^{\aleph_0}=2^{\aleph_0}$, the algebraic dimension of each $H\in\mathcal H$ is at most $2^{\aleph_0}$.

Next, we show that the infinite-dimensional separable Hilbert space $H_{\aleph_0}$ has algebraic dimension at least $2^{\aleph_0}$, by exhibiting a continuum of (algebraically) linearly independent elements in the space $\ell^2$ of square-summable sequences.

Let $(r_n:n\in\mathbb N)$ be an enumeration of the rational numbers. For $t\in\mathbb R$ and $n\in\mathbb N$ define $\varepsilon(t,n)$ to be $1$ if $r_n\lt t$ and $0$ otherwise. Finally, let $x_t=\left<\dfrac{\varepsilon(t,n)}n:n\in\mathbb N\right>\in\ell^2$. It is easy to see that every finite subset of $\{x_t:t\in\mathbb R\}$ is linearly independent.