# How to prove that any Hamel basis of an infinite-dimensional separable real Hilbert space is uncountable?

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

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?

• OP's space is not supposed to be complete, so how do you use Baire's category theorem ? – Olórin Jan 21 '15 at 13:25
• @RobertGreen : I have said it IS complete , read the first line – user123733 Jan 21 '15 at 13:26

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

• I never said anything about orthonormal – user123733 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 " ) – user123733 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 math.stackexchange.com/questions/217516/… – user123733 Jan 21 '15 at 13:20