Let $X$ be an inner product space.$M\subset X$. Prove that
a) $(span(M))^\bot=M^\bot$
b) $(\overline{M})^\bot=M^\bot$
My Work and problems:
a) Clearly $(span(M))^\bot\subset M^\bot$. Now let $x\in M^\bot$. Then $<x,y>=0$ for all $y\in M$. Let $z\in span(M)$. Then can I write $z$ as a finite linear combination of elements of $M$ even though $M$ is an infinite set?
b) Clearly $(\overline{M})^\bot\subset M^\bot$. How can I show the other inclusion?
Any help is appreciated. Thanks.
Regarding a) In fact, the definition of span of a set is the set of finite linear combination of element of respective set, i.e. $$ {\rm span}(M):= \{\sum_{i=1}^n c_i m_i | n\in {\mathbb N}, m_1,\ldots m_n\in M \& c_1,\ldots, c_n\in {\mathbb F}, \} $$ So you can do as you have said.
Regarding b) One may discuss by taking an arbitrary element $x$ in $M^{\perp}$. To show the claim, $x\in {\bar{M}}^{\perp}$, it is enough to prove that $<x,m>=0$, for any $m\in \bar{M}$. Note that for any $m\in\bar{M}$ there exists a sequence of elements of $M$ like $\{m_n\}_n$ such that $m_n\to m$. Thus, from continuity of inner product, it yields $$ <x,m> = <x,\lim_{n\to \infty} m_n> = \lim_{n\to \infty} <x, m_n> = \lim_{n\to \infty} 0 =0. $$