# Dimension of a set and its closure are equal in an Inner-product space?

I want to show that, given a subset $M$ of an Inner Product space $X$. If $M$ is a total set then, $M^\perp=\{0\}$. Which I have shown using the completion of $X$, which will be a Hilbert Space. And if a $A$ is dense in $X$ and $X$ is dense in $H$ then $A$ is dense in $H$.

I am trying another approach, where I clain $dim(V) = dim(\overline{V})$. (I've taken $V$ to be span of $M$). Is this claim always true? (I was posting my proof but I just found a flaw in it.)

-
What does dimension mean here? It's certainly not true of the usual Hamel dimension. –  Chris Eagle Apr 25 '12 at 15:13
I'm afraid I meant Hamel dimensions. –  swair Apr 25 '12 at 15:16

If you take the dimension of a vector space to be the cardinality of a Hamel basis of the space, your claim is false. For example, in $\ell_2$, let $V$ be the set of finite linear combinations of the usual unit vectors in $\ell_2$. Then the set of unit vectors is a Hamel basis of $V$. But the closure of $V$ is all of $\ell_2$ and it has been shown that a Hamel basis of a complete infinite dimensional normed space must be uncountable. See here, for example.
Right. Any Good way to prove the Lemma then? Namely, if M is total then $M^\perp =\{0\}$ –  swair Apr 25 '12 at 15:19