Question on a property of the orthogonal complement: $A\subset(A^\perp)^\perp$ Among the principal properties of the orthogonal complement, we have the following:
$$A\subset(A^\perp)^\perp$$
Where $A$ is a subset of an inner product space $X$, and $A^\perp$ is the orthogonal complement of $A$.
The proof given in Rynne and Youngson, "Linear Functional Analysis", is as follows:
"Let $a\in A$. Then for all $x\in A^\perp$, $(a,x)=\overline{(x,a)}=0$, so $a\in(A^\perp)^\perp$. Thus, $A\subset(A^\perp)^\perp$."
How is it, exactly, that this shows that $a\in(A^\perp)^\perp$? And consequently that $A\subset(A^\perp)^\perp$? Has it to do with the sesquilinear nature of the inner product, and that the second arguement is conjugate linear?
Furthermore, why is it that we do not have equality, namely that $A=(A^\perp)^\perp$? Clearly it is not the case, but when I was reading it and thinking about it, I thought that they should be equal.
 A: First, refresh some definitions:
Definition 1:

By definition, if $X$ is a subset of $V$, then $X^\bot$ is defined as $$X^\bot = \{v\in V: \forall x\in X:\langle x, v\rangle = 0\}$$

Definition 2:

By definition, $A$ is a subset of $B$ if and only if every element of $A$ is also an element of $B$, i.e. $$A\subset B\iff \forall x\in A: x\in B$$


Now, you want to prove that $A\subset (A^\bot)^\bot$. By definition 2, you must prove that every $a\in A$ is also an element of $(A^\bot)^\bot$.
So, you take an arbitrary $a\in A$. And you prove that for all $x\in A^\bot$, you have $\langle a, x\rangle = 0$. This, by definition 1, means that $a\in (A^\bot)^\bot$!

As far as your second question is concerned, it does sometimes happen that $A=(A^\bot)^bot$. For example, if $V$ is a vector space, then $(\{0\}^\bot)^\bot = \{0\}$ and $(V^\bot)^\bot = V$. Also, if $V$ is finite dimensional, you have equality. But in general, you do not.
A: To your second point:
In a Banach space it holds:
$$A\subseteq E:\quad\overline{\langle A\rangle}=\left(A^\perp\right)^\perp$$
(This fails in incomplete spaces.)
