Is there a subset $M$ of inner product space $U$ such that $(M^{\bot})^{\bot}$ is not the minimal closed subspace that contains $M$?

Question

Is there a subset $M$ of inner product space $U$ such that $(M^{\bot})^{\bot}$ is not the minimal closed subspace that contains $M$?

I have learned that when $U$ is Hilbert space there is no such a counter example. But what if $U$ is not Hilbert?

Answer 1

Let $U$ be the set of all $x \in \ell^2(\mathbb{N})$ for which $x = \{ x_n \}_{n=1}^{\infty}$ has only finite many non-zero elements. Then $U$ is an inner product space with the inner product inherited from $\ell^2(\mathbb{N})$. Define $$ M = \left\{ u \in U : \sum_{n=1}^{\infty}\frac{1}{n}u_n=0\right\}. $$ Then $M$ is closed in $U$ because it is the null space of a continuous linear functional $F(u)=\sum_{n=1}^{\infty}\frac{1}{n}u_n$ on $U$. And $M$ contains the following $$ \{ 1,-2,0,0,0,\cdots \} \\ \{ 1,0,-3,0,0,\cdots \} \\ \{ 1,0,0,-4,0,\cdots \} \\ \cdots. $$ So $M^{\perp}=\{0\}$. To see this, suppose $a=\{ a_1,a_2,a_3,\cdots \} \in M^{\perp}$. Then $$ a_2=\frac{1}{2}a_1,\;\; a_3=\frac{1}{3}a_1,\;\ a_4=\frac{1}{4}a_1, \cdots . $$ Because $a\in U$ must be true, then only finitely many $a_j$ are non-zero, which implies that $a_1=0$ and, hence, $a=0$. Therefore $(M^{\perp})^{\perp}=\{0\}^{\perp}=U$, but the closure of $M$ is not $U$.