# Can any infinite dimensional Hilbert space have countably infinite dimension?

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.

• It is so. You are missing nothing ! – Fred Sep 15 '20 at 6:59
• @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 $n \to \infty.$" – Anacardium Sep 15 '20 at 7:08

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.
• 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? – Anacardium Sep 15 '20 at 7:23