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Can any infinite dimensional Hilbert space have countably infinite dimension?

I know that there doesn't exist any infinite dimensional Banach space whose dimension is countably infinite for otherwise it would violate Baire's category theorem. What about infinite dimensional Hilbert spaces? Since Hilbert spaces are also Banach spaces the same result has to be true for Hilbert spaces. Isn't it so? Am I missing something?

Any suggestion regarding this will be highly appreciated. Thanks in advance.

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    $\begingroup$ It is so. You are missing nothing ! $\endgroup$ – Fred Sep 15 '20 at 6:59
  • $\begingroup$ @Fred I recently came across a problem which states that "Let $\{e_n \}_{n \in \Bbb N}$ be an orthonormal basis of a Hilbert space $H$ and $P_n$ the orthogonal projection onto $\text {span}\ \{e_1,e_2, \cdots ,e_n\},\ n \geq 1.$ Prove that for all bounded linear operator $T: \mathcal H \longrightarrow \mathcal H$ and $h \in H,$ $P_nTP_nh \to Th$ as $n \to \infty.$" $\endgroup$ – Anacardium Sep 15 '20 at 7:08
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Yes you are correct. But note that we are talking about a Hamel basis here. Every Hilbert space also has a more useful notion of basis, called orthonormal basis, which is a set of orthonormal elements in the Hilbert space with dense span. Such a basis can be countable, for example $l^2(\Bbb{N})$ has orthonormal basis the Dirac functions on the natural numbers.

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  • $\begingroup$ I haven't found it anywhere in the book Functional Analysis by Walter Rudin. Can you please share me some references? $\endgroup$ – Anacardium Sep 15 '20 at 7:21
  • $\begingroup$ From my knowledge of vector space I know that if $\mathcal B$ is a basis of an infinite dimensional $K$-vector space $V$ then every element of $V$ can be written as a finite $K$-linear combinations of vectors in $\mathcal B.$ Is it wrong then? $\endgroup$ – Anacardium Sep 15 '20 at 7:23
  • $\begingroup$ Here are some references from the top of my head: Axler's measure theory, Folland's real analysis, Conway's functional analysis. Probably Rudin's book real and complex analysis has it also. The difference is that with an orthonormal basis you need limits of finite sums. $\endgroup$ – QuantumSpace Sep 15 '20 at 13:07
  • $\begingroup$ @Anacardium Is something still unclear? $\endgroup$ – QuantumSpace Sep 17 '20 at 16:17
  • $\begingroup$ Do you want your answer to be accepted for getting a benefit of 10 reputation points? $\endgroup$ – Anacardium Sep 17 '20 at 16:31

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