Orthogonal complement of vector spaces Let $V$ be a vector space. Here I do not restrict $V$ to be finite dimensional. Let $S$ be a vector subspace of $V$. Why is $S\subset (S^{\perp})^{\perp}$ rather than $S= (S^{\perp})^{\perp}$? 
 A: Well, first off, we actually need to assume $V$ is an inner product space with fixed inner product to make sense of orthogonality. Now, an inner product induces a norm, and from this a topology. With respect to this, the orthogonal complement of any set is a closed vector subspace of $V$. So obviously, if $S$ isn't closed in the topology given by the inner product, it cannot be equal to $(S^\perp)^\perp$.
Example: Let $V = \ell_2(\mathbb{N})$ be the space of all sequences of real numbers (I know...I know, complex) that are square-summable, that is $\{a_n\}_{n\in\mathbb{N}}$ such that 
$$\sum_{n=1}^\infty a_n^2 < \infty$$
This has an inner product:
$$\langle \{a_n\},\{b_n\}\rangle = \sum_{n=1}^\infty a_nb_n$$
Now consider the subspace of all sequences with only finitely many non-zero entries as $S$. For all $n$, $S$ contains the sequence with a $1$ in the $n$-th place and no other non-zero entries, call this $e_n$. If a sequence $a \perp e_n$ then $a_n = \langle a,e_n\rangle= 0$. So if $a \perp S$ then $a_n = 0$ for all $n$. So $S^\perp$ is just the zero vector, and $(S^\perp)^\perp = V \neq S$. 
A worthwhile exercise is to try and prove that this (failure to be closed) is basically the only thing that can go wrong in the sense that $(S^\perp)^\perp = \overline{S}$ in a complete inner product space (that is, a Hilbert space)
