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Hilbert Space is an "infinity" dimensional vector space. Does the "infinity" means $\aleph^0$ or $\aleph^1$ ? Or it does not matter at all?

Math newbie thanks you.

Could you please up vote for once so I could comment on others' posts?

High GPA
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    A Hilbert space can be finite-dimensional. Take the euclidean space $\mathbb{R}^n$ for example. – Augustin May 17 '17 at 16:06
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    Notably, any infinite-dimensional Hilbert space is necessarily of uncountable dimension, if dimension is defined in the sense of Hamel bases. – Ben Grossmann May 17 '17 at 16:09
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    Also, if you want more reputation, you should try either answering questions or editing the incoming posts. – Ben Grossmann May 17 '17 at 16:12
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    @Omnomnomnom Reputation looks like a fun stuff but I am more interested in learning things that I didn't understand well. – High GPA May 17 '17 at 17:30
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    @LowGPA sure, but site privileges are "privileges" for a reason; you shouldn't expect things to be as convenient as possible for you just because you're "interested in learning things". In any case, it looks like you got those points you were after. – Ben Grossmann May 17 '17 at 17:38
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    It depends whether you allow infinite linear combinations or not. See https://en.wikipedia.org/wiki/Basis_(linear_algebra)#Analysis. – tparker Jun 18 '17 at 03:40

3 Answers3

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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.

Ethan Bolker
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  • In this case, is it correct to say that there's only one countably infinite dimensional Hilbert space? Since any Hilbert space is isometric to $\ell_{2}$? Been trying to wrap my head around that. – Gustavo De Souza Mar 24 '24 at 17:35
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    Yes, there is (up to isometric isomorphism) just one Hilbert space of each dimension - an orthonormal basis in each defines an automorphism. – Ethan Bolker Mar 24 '24 at 18:16
  • Got it. This dimension is the Hilbert dimension, that is, the cardinality of the orthonormal basis, correct? I'm only double checking because this really confused me, since I was only familiar with finite dimensional vector spaces – Gustavo De Souza Mar 24 '24 at 19:01
  • Yes. (Previous comment says automorphism instead of isomorphism.) – Ethan Bolker Mar 24 '24 at 19:29
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A Hilbert space need not be infinite-dimensional as tilper observed. However, if a Hilbert space is infinite-dimensional, then it is uncountable-dimensional; in fact, it has dimension at least $2^{\aleph_0}$. Incidentally, it turns out that this may be strictly bigger than $\aleph_1$!

Noah Schweber
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  • Why does it have to be at least $2^{\aleph_0}$ ? – High GPA May 17 '17 at 16:30
  • @LowGPA My previous phrasing wasn't too clear - I've rewritten the answer to better separate the two links (and linked to the relevant answer, not just the question). – Noah Schweber May 17 '17 at 16:33
  • I am a little bit confused about your last sentence. Is the infinte number of dimensions of Hilbert space strictly "bigger than" or "bigger than or equal to" ${\aleph}_1$ ? – High GPA May 17 '17 at 19:31
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    @LowGPA Well, it's certainly $\ge\aleph_1$ - we can prove that $\aleph_1\le 2^{\aleph_0}$. The issue is that the statement "$\aleph_1=2^{\aleph_0}$" (called the "continuum hypothesis") isn't provable or disprovable from the usual axioms of set theory. Note that I'm not saying "no proof/disproof is known," but rather "we can prove that no proof/disproof exists (unless ZFC is inconsistent)"! So there's a fundamental barrier here. Here, the relevant thing we can prove is that the dimension must be at least $2^{\aleph_0}$. – Noah Schweber May 17 '17 at 19:35
  • I finally understand what you mean in the strict math sense. Thanks – High GPA May 17 '17 at 19:37
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A Hilbert space is not necessarily infinite dimensional.

A Hilbert space is an inner product space that's complete with respect to its norm. For example, $\Bbb R^3$ with the usual Euclidean norm and dot product is a Hilbert space of dimension $3$.