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According to Wikipedia,

Hilbert space [...] extends the methods of vector algebra and calculus from the two-dimensional Euclidean plane and three-dimensional space to spaces with any finite or infinite number of dimensions

However, the article on Euclidean space states already refers to

the n-dimensional Euclidean space.

This would imply that Hilbert space and Euclidean space are synonymous, which seems silly.

What exactly is the difference between Hilbert space and Euclidean space? What would be an example of a non-Euclidean Hilbert space?

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    $\begingroup$ "Euclidean space" means simply "finite dimensional Hilbert space". Nothing more nothing less. So, Hilbert space is more general than Euclidean space, since a Hilbert space may not be finite dimensional. $\endgroup$ – Crostul Aug 25 '16 at 13:28
  • $\begingroup$ @Crostul That's quite clear now! $\endgroup$ – Yu Gu Sep 9 '18 at 12:06
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A Hilbert space essentially is also a generalization of Euclidean spaces with infinite dimension.


Note: this answer is just to give an intuitive idea of this generalization, and to consider infinite-dimensional spaces with a scalar product that they are complete with respect to metric induced by the norm. Clearly, there are finite-dimensional Hilbert spaces, as $\mathbb{R}^n$, with the standard scalar product and Euclidean metric.

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Hilbert space: a vector space together with an inner product, which is a Banach space with respect to the norm induced by the inner product

Euclidean space: a subset of $\mathbb R^n$ for some whole number $n$

A non-euclidean Hilbert space: $\ell_2(\mathbb R)$, the space of square summable real sequences, with the inner product $((x_n),(y_n)) = \sum_{n=1}^{\infty}x_n y_n$

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    $\begingroup$ Not a 'subset' of $\mathbb{R}^n$. Nor necessarily is a Euclidean space $\mathbb{R}^n$ for any $n$: for instance, the space of linear functionals on $\mathbb{R}^n$ is not $\mathbb{R}^n$, although it is naturally isomorphic to it, as indeed any Euclidean space is isomorphic to $\mathbb{R}^n$, once a basis has been specified. But it is better to be basis-free as much as possible. $\endgroup$ – Calum Gilhooley Aug 25 '16 at 13:49
  • $\begingroup$ @CalumGilhooley May I ask you what is the definition of Euclidean space(I'm confused by what wiki said)? Does it necessarily have a inner-product structure(i.e. Euclidean space is necessarily a inner-product space)? Or it just something that satisfy some "geometric" axioms, such like Euclid's or Hilbert's? $\endgroup$ – Eric Jan 15 '17 at 4:38
  • $\begingroup$ @Eric I confess to also being confused by the term "Euclidean space", and the Wikipedia article does not resolve my confusion. I think the most common use of the term, and also the clearest, is that given by Crostul's comment on the OP, viz. a Euclidean space is any finite-dimensional real inner product space. (All such spaces are complete, i.e. are Hilbert spaces.) But the definition of the term in this article is more restrictive. So I think one has to be careful, and look to the context, and define the term if using it oneself. $\endgroup$ – Calum Gilhooley Jan 16 '17 at 16:24
  • $\begingroup$ @CalumGilhooley Thanks. So how the Hilbert's or Euclid's axioms come into play in modern geometry? Logically speaking, if we choose the definition of a euclidean space to be real inner product space, meaning that we don't follow/need their(Euclid's or Hilbert's) axiom anymore right? For example, there's a Hilbert's axiom "For every two points A, B there exists a line a that contains each of the points A, B." but in inner-product space terminology, it can be directly proved by trivial pre-calculus techique. $\endgroup$ – Eric Jan 16 '17 at 17:13
  • $\begingroup$ I also asked a new post here, maybe you can move there to help me. :) $\endgroup$ – Eric Jan 16 '17 at 17:15

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