I got the following question:
If $N$ is a linear subspace of the dual space $X^{\prime}$, the annihilator of $N$ is defined as $$ N^{\perp}:=\{x \in X: f(x)=0 \text { for all } f \in N\} \subset X \text {. } $$
$$ M^{\perp}:=\left\{f \in X^{\prime}: f(y)=0 \text { for all } y \in M\right\} \subset X^{\prime} \text {. } $$
Prove: If $X$ is a Banach space, then $\bar{N} \subset\left(N^{\perp}\right)^{\perp}$.
My attempt:
$N^\perp$ is the set of all points $x \in X$ on which all $f \in N$ vanish.
$(N^\perp)^\perp$ is the set of all $g \in X^\prime$ that vanish on $N^\perp$
$\Rightarrow N \subset (N^\perp)^\perp$
Let $f_n$ be a sequence in $N$ with limit point $f$ (in the operator norm)
Then $|f_n(x) - f(x)| = |(f_n-f)(x)| \leq ||f-f_n|| \; ||x||$
$||f-f_n|| \rightarrow 0 \Rightarrow |f_n(x) - f(x)| \rightarrow 0$
$\Rightarrow f_n(x)$ converges to $f(x)$ point wise
Let $x \in N^\perp$:
$f(x) = \lim_{n \rightarrow\infty} f_n(x) = \lim_{n \rightarrow\infty} 0 = 0$
$\Rightarrow \overline{N} \subset (N^\perp)^\perp$
But I am really not sure if this is correct because I don't think I used the assumption that $X$ is a Banach space anywhere... So I wanted to ask if I'm missing something
