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How to prove that any Hamel basis of an infinite-dimensional complete and separable (having a countable dense set ) real inner-product space is uncountable ? Do I have to use Baire-category theorem ? Please help

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Since the space is infinite dimensional so its basis cant be finite.Now every complete metric space is of $2nd$ category and by Baire Category theorem it cant be expressed as a countable union of Nowhere dense sets .can you take it from here?

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The open balls of radius $\frac{1}{2\sqrt2}$ around every point belonging the the orthonormal basis vectors are disjoint. A countable set can't intersect them all if there are uncountably many vector in you Hamel basis, so it is not dense, and then the space is not separable. So there is a problem with your set of hypothesis.

What preceeds is conditional to the fact that one can orthonormalize the Hamel basis, which we can :

Let me note $E$ your space. For a family $\{\,e_j\mid j\in J\,\}\subseteq E$ note $s(w)$ be the space of all $\sum_{j\in J}x_je_j$ with $\sum \lvert x_j e_j\rvert^2<\infty$ and with the set of $j$'s such $x_j\ne 0$ countable.

Call a "finite Gram-Schmidt" as a subset $J\subseteq I$ together with a total order $\le$ on $J$ and an orthonormal family $\{\,y_j\mid j\in J\,\}$ such that $y_j\in s(v)\cup\{\,y_k\mid k\in J, k<j\,\})$ for each $j\in J$.

The set of finite Gram-Schmidt is inductively ordered. Then Zorn's lemma ensures that this set a maximal element for the order. We must have $J=I$. (Do you see why ?)

We do have a Gram-Schmidt orthonormalization process in $E$. If we apply it to your Hamel base, we can suppose you Hamel basis to be orthogonal

Olórin
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  • I never said anything about orthonormal –  Jan 21 '15 at 12:58
  • But doesn't Gram-Schmidt orthogonalization process work in infinite dimensional spaces ?... – Olórin Jan 21 '15 at 13:00
  • I think not , in fact that is what I am willing to prove that it does not work for Infinite dimensional separable Hilbert spaces ( where by basis I mean "Hamel Basis " ) –  Jan 21 '15 at 13:02
  • @user123733 I edit my answer, showing we can orthonormalize the Hamel basis. – Olórin Jan 21 '15 at 13:16
  • Is a Hilbert space also a Banach space ( I think yes ) ? see this http://math.stackexchange.com/questions/217516/let-x-be-an-infinite-dimensional-banach-space-prove-that-every-hamel-basis-of –  Jan 21 '15 at 13:20
  • @user123733 Yes, by definition, a Hilbert space is vector space with an inner product such that the space is complete wrt norm induced by the inner product. In terms of topology, you will note the conceptual similarity between the 1st paragraph of my answer and the proof in the link you just gave ;-) – Olórin Jan 21 '15 at 13:22
  • @user123733 My proof works without use of being complete ;-) – Olórin Jan 21 '15 at 13:24
  • A complete real inner-product space is a Hilbert space , that's what my question says ; and what conceptual similarity ; you have claimed the opposite –  Jan 21 '15 at 13:26
  • @user123733 sorry, haven't seen complete, by bad ! – Olórin Jan 21 '15 at 13:26
  • I doubt that you can make it work without completeness. A metric space is separable iff it is second countable, which is inherited by subspaces. Take $(X, \langle -, - \rangle)$ a separable Hilbertspace and $\mathscr{B} = {e_i}{i \in \mathbb{N}}$ be an infinite linearly independent set in $X$. Then the subspace $(X', \langle-|{X'},|_{X'}\rangle)$ of $(X, \langle -, - \rangle)$ (algebraically) spanned by $\mathscr{B}$ is an infinite-dimensional (not complete) separable real inner-product space with countable basis. – Carlo K. Nov 06 '21 at 10:23
  • -1: this answer is wrong. – Martin Argerami Jul 11 '22 at 21:39