Computing the orthogonal complement of a subset (dimension p) of a finite dimensional vector space (of dimension $n$) is quite straight forward.
Simply find a basis, make it orthonormal, construct the orthogonal projection and then choose n-p vectors and calculate their orthogonal projection. Then said vectors minus their orthogonal projections will be in the orthogonal complement. However I'm not quite sure how to do something like this in the infinite dimensional case. In particular this is the Vector space, and subspace I am working with: $$ V=\{ (a_0, a_1, a_2, ...)\vert \sum_{n=0}^{\infty}\vert a_n \vert ^2 < \infty \}$$ $$ \langle(a_n)_{n=0}^{\infty},(b_n)_{n=0}^{\infty}\rangle := \sum_{n=0}^{\infty}a_n \cdot b_n $$ $$ U = \{ (a_n)_{n=0}^{\infty} \in V \vert \exists N \geq 0 \text{ s.t. } \forall m \geq N : a_m = 0\} $$ I'm not quite sure how to approach this, so far I have: $$ \forall (a_n)_n \in U, (b_n)_n \in U^\perp : \langle(a_n)_n, (b_n)_n\rangle = 0 \Longleftrightarrow \sum_{n=0}^N a_n \cdot b_n = 0 $$ So basically the product of a sequence in $U$ and any other one, ends up in $U$ again. I don't think this helps me though. Any ideas on how to continue?
$U^{\perp}=\{b\in V: \langle a, b\rangle=0\}$
$\begin{align}\langle a, b\rangle&=\sum_{n\ge 0}a_n b_n\end{align}$
$\forall i\in\Bbb{N} ,e_i=(0, 0,\ldots , 1,0,0\ldots)\in U $
Hence $\langle e_i, b\rangle=b_i=0$
Hence $b=(b_i) =(0, 0,\ldots) $
$U^{\perp}=\{0\}$
Note: $V$, $U$ has special notation : $\ell_2$ and $\textrm{c}_{00}$ respectively.