Orthogonal complements of subsets I wanted to show that $U\subset W \implies W^\perp \subset U^\perp$. I have managed to show the implication that $U\subset W \implies W^\perp \subseteq U^\perp$ and I'm not quite sure how to make the subset strict?
 A: The claim, as written, is not true. Consider the vector space $\mathbb{R}^3$ and let $W = \mbox{span}\{(1,0,0),(0,1,0)\}$ and $U=W\backslash\{(0,1,0)\}$, i.e. $U$ is a the plane $W$ without one point. Then, $U\subset W$ but $W^\perp=U^\perp = \mbox{span}\{(0,0,1)\}$.
Consider instead a vector space $\mathcal{X}$ with $U\subset W\subset \mathcal{X}$ and $U$ and $W$ are linear subspaces. It is straightforward to show that $W^\perp\subseteq U^\perp$. Then, we can write $\mathcal{X} = U\cup U^\perp$. Let $w\in W\backslash U$, which must exist since $U\subset W$. Then, $w\in U^\perp$ because $\mathcal{X}=U\cup U^\perp$ but also $w\not\in W^\perp$ since $w\in W$ so in fact we have $W^\perp\subset U^\perp$.
A: As Tony points out in his answer, the claim is not true if $U$ is not a linear subspace, since the orthogonal complement $U^\perp$ is orthogonal to the linear closure of $U$.
Similarly, if $U$ is a linear subspace of a topological vector space, but is not a closed set, then we may have for example $U\subsetneq \bar U$, but then $U^\perp = \bar U^\perp.$
For example, in $\ell^2(\mathbb{N}),$ let $U=c_0$ be the sequences which are eventually zero. Then $U\subsetneq \ell^2(\mathbb{N})$, but $U^\perp= \ell^2(\mathbb{N})^\perp = 0.$
