Orthogonal complement for infinite dimensional and open vector spaces.

Question

Given a vector space $V$ and a subspace $U$, the orthogonal complement $U^{\bot}$ has the following properties:

1. $U^{\bot}$ is itself a subspace of V.
2. $U \cap U^{\bot} = \{0\}$.

3. $(U^{\bot})^{\bot} \supseteq U$.

   3'. $(U^{\bot})^{\bot} = U$ if $U$ is a closed subspace (All Cauchy's sequences converge and $U$ contains all their limit points)

Can you please give me an (easy) example of a vector space $V$ and a subspace $U$ where $(U^{\bot})^{\bot} \supset U$ and show me that there is in fact a vector which is in $(U^{\bot})^{\bot}$ but NOT in $U$?

Answer 1

We need infinite dimensional vector spaces to produce a counterexample.

Consider the subspace of $V = \ell^2$ consisting of sequences $(x_n)$ such that all but finitely many $x_n$ are zero. Take this to be our $U$.