Take the Hilbert space $\ell^2$ of all $x=(x_1,x_2,...x_i...)$ such that $x_i \in \mathbb R$ for all $n$, with $\sum_{i=1}^∞ x_i^2<∞$ and the defined inner product $<x,y> = \sum_{i=1}^∞x_iy_i$.
Take a subset of this space $S$. Let such $S$ be the set of all elements $x=(x_1,x_2,...x_i...)$ with the condition that $x_{2n} = 0$ for $n\ge1$. Then what is the orthogonal complement of $S$?
So far, clearly all such $x_{2n}$ are in the orthogonal complement, since the inner product of these elements with any element of $S$ is zero. $0$ too is of course in $S^⊥$. Where do I go from here? It's fine to say what $S^⊥$ 'is', but how do I actually completely define it?
You have a Hilbert base for $S$ namely $X=\{e_{2n-1}|n\geqslant 1\}$ (that is, $S$ is the closure of the linear subspace spanned by $X$), where $e_{k}=(e_{k,i})_{i}$ is the vector having $e_{k,i}=0$ for $i\neq k$ and $e_{k,k}=1$. If $X$ is a dense subset of $S$, then $$S^\perp=X^\perp=\{z|\langle z,x\rangle=0\forall x\in X\}$$ (clearly $S^\perp\subseteq X^\perp$, while $X^\perp\subseteq S^\perp$ because every $s\in S$ is the limit of a sequence $x_n$ in $X$ and so $z\in X^\perp\rightarrow\langle z,s\rangle=\langle z,\lim_nx_n\rangle=\lim_n\langle z,x_n\rangle=0\rightarrow z\in S^\perp$).
In this case the dense subset is $\{e_{2n-1}|n\geqslant1\}$, so $$S^\perp=\{z|\langle z,e_{2n-1}\rangle=0\forall n\geqslant 1\}=\{z|z_{2n-1}=0\forall n\geqslant 1\}$$