Hilbert Space is an "infinity" dimensional vector space. Does the "infinity" means $\aleph^0$ or $\aleph^1$ ? Or it does not matter at all?
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[Math] Is the number of dimensions in Hilbert Space countable infinity or uncountable infinity
functional-analysishilbert-spacesinfinity
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[Math] the difference between a complete orthonormal set and an orthonormal basis in a Hilbert space
For any infinite sequence $u_n$ of distinct members of your orthonormal set, you can take convergent infinite sums such as $\sum_{n=1}^\infty u_n/n$, i.e. $\lim_{N \to \infty} \sum_{n=1}^N u_n/n$ (convergent in the Hilbert space norm). The limit exists because the Hilbert space is a complete metric space. It's easy to prove that the limit is not a linear combination of finitely many members of the orthonormal set.
A complete orthonormal set in a Hilbert space is called an "orthonormal basis", but this use of the term "basis" is different from the ordinary vector space "basis".
"Copy of X" here means "space that is isomorphic to X". So the condition is indeed that $H$ should contain infinitely many subspaces, each of which is isomorphic to the orthogonal complement $H^\perp$, and which are pairwise orthogonal to each other.
A concrete example might be $K = L^2(\mathbb{R})$, $H = L^2(0,\infty)$. Then $H^\perp = L^2(-\infty,0)$. Let $H_i = L^2(i, i+1)$, which you can show is isomorphic to $L^2(-\infty, 0)$, and the $H_i$ are all pairwise orthogonal. So $H$ is an example of a large subspace.
The notation $K \setminus H$ here doesn't mean set difference, but orthogonal complement. So $K \setminus H$ is just the orthogonal complement of $H$ in $K$, which you could also denote $H^\perp$ when the ambient space $K$ is understood. It is certainly a closed subspace of $K$, and it makes sense to speak of its dimension.
The $K \setminus H$ notation is fairly common, and although it is technically ambiguous with set difference, in practice it is usually clear from context which one is meant.
Let $H$ be a subspace of the separable Hilbert space $K$. It is clear that a large subspace must be infinite-dimensional (it contains infinitely many nontrivial orthogonal subspaces).
For the converse, suppose $H$ is infinite-dimensional, and fix an orthonormal basis $\{e_1, e_2, \dots\}$ for $H$. Let $H^\perp$ be its orthogonal complement.
Suppose first that $H^\perp$ has finite dimension $n$. For each $i > 0$ let $H_i$ be the subspace of $H$ spanned by $\{e_{ni+1}, \dots, e_{ni+n}\}$. Each $H_i$ is isomorphic to $H^\perp$ because they have the same dimension $n$, and the $H_i$ are pairwise orthogonal because they are spanned by disjoint subsets of an orthogonal set. Thus $H$ is large.
Otherwise, suppose $H^\perp$ is infinite-dimensional; since it is separable, it has a countable orthonormal basis $\{f_1, f_2, \dots\}$. Let $\{ k_{1,1}, k_{1,2}, \dots\}$, $\{k_{2,1}, k_{2,2}, \dots\}$ be infinitely many disjoint sequences in $\mathbb{N}$; for instance you could enumerate the primes as $p_n$ and let $k_{n,i} = p_n^i$. Let $H_i$ be the closed linear span of $\{e_{k_{i,1}}, e_{k_{i,2}}, \dots\}$. Each $H_i$ is isomorphic to $H^\perp$, because $H^\perp, H_i$ are both infinite-dimensional separable Hilbert spaces, or because they both have orthonormal bases of the same cardinality $\aleph_0$. And the $H_i$ are again pairwise orthogonal because they are the closed linear spans of disjoint orthogonal sets. Thus $H$ is also large in this case.
In essence, it's really just the fact that a countably infinite set can be written as a countable union of countable sets (or of finite sets), i.e. that $\aleph_0 = \aleph_0^2$, applied to dimension instead of cardinality.
The term "large subspace" seems to have been introduced by Halmos and I haven't seen it used by anyone else; the concept is not one that I think comes up especially often. I wouldn't particularly expect there to be a lot of reference material on the topic. It does appear in Halmos's Hilbert Space Problem Book on page 131, where he says that "the idea has appeared before, even if the term has not".
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To clarify:
The Hilbert space of square summable sequences (the usual first one you encounter in analysis) does indeed have uncountable dimension when you're thinking about the cardinality of a basis such that every vector is a finite linear combination of basis elements.
But it's useful and routine to think of the countable set of sequences with just one nonzero entry that's $1$ as a basis (the standard basis) for purposes of analysis: every sequence is a limit in the Hilbert space topology of finite sums of those basis elements - i.e. a linear combination of (possibly) infinitely many of them.