A question related to $S^{\perp}$ and closure of span of $S$ This question was asked in my linear algebra quiz previous year exam and I was unable to solve it.

Let V be an inner ( in question it's written integer , but i think he means inner) product space and S be a subset of V. Let $\bar S$denote the closure of S in V with respect to topology induced by the metric given by inner product. Which of the following statements are  true?

A $ S $=$ (S^{\bot})^{\bot}$
B $ \overline S$= $(S^{\perp})^{\perp}$
C $\overline {\text{span}(S)}$=$(S^{\bot})^{\bot}$
D $ S^{\bot} $=$ ((S^{\bot})^{\bot})^{\bot}$
I was completely blank on how can I approach this problem although I have studied linear algebra carefully. Can you please tell on how I should approach the problem .
Edit : I tried It again . I marked A ,D but answer is C,D. If A is false I don't see why D must be true. So, I think I am missing some concepts.
 A: Recall that $S^\perp$ is defined as the set of all vectors in $V$ which are orthogonal to every vector in $S$.
In case $S$ is a
singleton, say $S=\{s\}$, then we have
$$
  S^\perp = \{x\in  V: \langle x, s\rangle =0\},
  $$
so   $S^\perp$ coincides with the null space of the continuous linear functional
$$
  x\in  V \mapsto  \langle x, s\rangle ,
  $$
and for that reason   $S^\perp$ is obviously a CLS (closed linear subspace).
For a general set $S$, one clearly has that
$$
  S^\perp = \bigcap_{x\in  S} \{s\}^\perp,
  $$
so $S^\perp$ is the intersection of a family of CLS's, and hence itself a CLS.
Notice that the right-hand-side of (A), (B), (C) and (D) all refer to the "perp" of something, so are all CLS's.
Since the left-hand-side of (A) and (B) may or may not represent a CLS, then they can't always be true.

Point (C) is true.  To see why,  first notice that
$$
  S\subseteq (S^\perp)^\perp
  \tag 1
  $$
for a pretty elementary reason (which
nevertheless sounds a bit like a tongue-twister):  every  vector in  $S$ is  orthogonal to anything that is orthogonal to every vector in $S$.
We then see that (1) states that $S$ is contained in a CLS, and since $\overline{\text{span}}(S)$ is the smallest CLS
containing $S$, it follows that
$$
  \overline{\text{span}}(S)\subseteq (S^\perp)^\perp
  $$
To prove the converse inclusion,  pick any vector  $x$ in  $(S^\perp)^\perp$.
A well known result about Hilbert spaces (which requires that $V$ be complete,  and hence  we need to assume it here)
states that $x$ may be written as
$$
  x=u+v,
  $$
where $u$ is perpendicular to $\overline{\text{span}}(S)$, and $v$ velongs to $\overline{\text{span}}(S)$.
Notice that, in particular,  $u$ is perpendicular to every vector of $S$, and hence $u\in  S^\perp$.
On the other hand,
since both $x$ and $v$ lie in $(S^\perp)^\perp$,  we conclude that $u=x-v\in (S^\perp)^\perp$.
This implies that $u\in  (S^\perp)^\perp \cap  S^\perp$,  so $u$ is perpendicular to itself,  whence $u=0$ and then
$$
  x = u+v = v\in  \overline{\text{span}}(S).
  $$
This concludes the proof of (C).

Regarding point (D) it is true even if $V$ is not complete.  It is a consequence of  the following  much more general result:
Lemma.  Let $V$ be any set (e.g. the inner-product space of interest here) and let $\lozenge$ be a symmetric relation on $V$
(e.g. $x\mathrel{\lozenge} y \Leftrightarrow x\perp y$).  For each subset $S\subseteq V$ define
$$
  S^\lozenge = \{x\in V: x\mathrel{\lozenge} s, \text{ for all $s$ in S} \}.
  $$
Then
$$
  S^\lozenge = ((S^\lozenge)^\lozenge)^\lozenge,
  $$
for any $S$.
Proof.  The tongue-twister above immediately implies that
$$
  S\subseteq (S^\lozenge)^\lozenge.
  \tag 2
  $$
Plugging in  $S^\lozenge$ in
place of $S$ in (2), we get $S^\lozenge \subseteq  ((S^\lozenge)^\lozenge)^\lozenge$.
Next observe that
$$
  S_1\subseteq S_2 \Rightarrow   S_2^\lozenge\subseteq S_1^\lozenge,
  $$
and if this is applied to (2), we get
$
  ((S^\lozenge)^\lozenge)^\lozenge\subseteq   S^\lozenge.
  $
QED
An interesting Corollary,  in a totally distinct area of Math is:
Corollary.   Given a ring $R$ and any subset $S\subseteq R$,  define the commutant of $S$, denoted $S'$, to be the set formed by
the elements of $R$
which commute with every element of $S$.    Then $S'''=S'$.
A: The condition that $S$ is a subset (and is not, for example, given to be a subspace of $V$) is rather odd, and makes some of this problem different from standard linear algebra. In general, in any inner product space, $S^\perp$ is a closed subspace, so the right sides of each of A, B, C, and D are closed subspaces.
However, for each of $A$, $B$, and $D$ (the original version of $D$, and not the current one), it is possible for the left side not to be a closed subspace. (The example given by Peter Franek of when $S$ is a one-element set is likely the simplest.) In general, though, $\operatorname{span}(S)$ should be a subspace, and $\overline{\operatorname{span}(S)}$ should be a closed subspace, so it should seem plausible that C could be true. It in fact is, as it essentially only deals with subspaces --- see if you can prove this.
A: $A,B,C$ are false, in general, and $D$ is true. If $V$ is Hilbert, then also $C$ is true.
Let $S$ be any set, then $S^\perp$ is the set of all vectors that are perpendicular to all elements of $S$. That's usually a definition, $S^\perp := \{v\,|\,\,\forall\,s\in S\,\,g(v,s)=0\}$.
You can easily verify that this is a vector subspace. Namely, if $v,w\in S^\perp$ and $\alpha\in \mathbb{R}$, then also $v + \alpha w\in S^\perp$.
So also $(S^\perp)^\perp$ is a vector subspace.
This immediately excludes $A$ and $B$, in cases when $S$ (or $\overline{S}$) is just a subset of $V$ but not a vector subspace. You can find plenty of counter-examples, for instance  $S=\{v\}$ for one vector $v\neq 0$.
$C$ is harder. The easy part is $LHS \subseteq RHS$.
Let $v$ be from the LHS. That is, $v$ is a limit of some $v_i$ such that $v_i\in \text{span}(S)$. Each $v_i$ is a combination $\beta_j s_j$ of elements of $S$. If $w\in S^\perp$, then $w$ is orthogonal to all elements of $S$ and hence $g(w, s_i)=0$ and also $g(w, v_i)=0$. Using continuity of $g$, we have $g(w, v) = g(w, \lim_i v_i) = \lim g(w, v_i) = 0$, so $w$ is also orthogonal to $v$. So $v$ is in $(S^\perp)^\perp$.
For the other implication, you need to show that a closed subspace (closure of the span of $S$ here) has an orthogonal complement. This is not always the case: there are proper closed subspaces $Y$ such that $Y^\perp = 0$!
So take such $Y$ as your $S$, then the span of $S$ is $S$ and the closure of $S$ is $S$, and $(S^\perp)^\perp = V \neq S$.
However, this being said, $C$ holds under some additional assumptions, such as $V$ being a Hilbert space.
If $C$ is true, then $D$ follows easily, because it reduces to
$$
S^\perp = (\overline{span(S)})^\perp
$$
which is obvious.
In general, if $V$ is not complete, then the proof of $D$ is given by the answer of @Ruy.
