0
$\begingroup$

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.)

  • 0
    I'm afraid I meant Hamel dimensions.2012-04-25

1 Answers 1

2

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.

  • 0
    Right. Any Good way to prove the Lemma then? Namely, if M is total then $M^\perp =\{0\}$2012-04-25