Show that $\overline{[S]}^{\bot}=S^{\bot}$, for any subset of a inner product space.

Question

I have done a solution for this problem:

Let $X$ be an inner product space and $S$ a subset of $X$. Show that $\overline{[S]}^{\bot}=S^{\bot}$.

But I am note sure that my solution is correct. If someone could help me and look at my solution I would be pleased.

My solution is:

As $\overline{[S]}\supset S \Rightarrow \overline{[S]}^{\bot}\subset S^{\bot}$. Now as $S\subset S^{\bot\bot}\Rightarrow [S]\subset [S^{\bot\bot}]=S^{\bot\bot}$ moreover $$\overline{[S]}\subset\overline{S^{\bot\bot}}=S^{\bot\bot}\Rightarrow \overline{[S]}^{\bot}\supset S^{\bot\bot\bot}\supset S^{\bot}$$ which finishes the solution. $\square$

Answer 1

Your solution looks fine. If you have doubts, here's an approach which doesn't use double and triple complements:

As you have pointed out, from $S \subset \overline{[S]}$ follows $S^\perp \supset \overline{[S]}^\perp$. We still need to show the other inclusion.

Let $x\in S^\perp$ and $y \in \overline{[S]}$. Then there exist $y_n \in S$ such that $y_n \rightarrow y.$ We can write each $y_n$ as a linear combination of vectors in $S$: $y_n= \sum_{i=0}^{K_n} \alpha^{(n)}_i s_i $ We have:

$$\langle x,y \rangle = \lim_{n\rightarrow \infty} \langle x , y_n \rangle = 0,$$ where the last equality follows from $\langle x,y_n \rangle = \langle x, \sum_{i=0}^{K_n} \alpha^{(n)}_i s_i \rangle = \sum_{i=0}^{K_n} \alpha^{(n)}_i \langle x, s_i \rangle = 0$.