Proving that $S=\{ x \in (X,\| \cdot \|) : \|x\| =1 \}$ is a closed set. Exercise :

Show that the unit sphere $$S=\{x \in (X,\|\cdot \|) : \|x\| =1\}$$ of a normed space, is a closed set.

Attempt :
For a set to be closed, its complement must be an open set. Define the complement of $S$ to be :
$$S^c = \{ x \in (X, \|\cdot \|) : \|x\| <1 \}\cup\{x\in (X,\|\cdot\|):\|x\|>1\}$$
Note : We have NOT yet been introduced to handling maps for such proofs in our functional analysis course, so it may not be the best approach for my understanding.
Question : How would one proceed with proving the fact stated in the exercise?
 A: You cannot define freely the complement of a set. In this case,$$S^\complement=\bigl\{x\in X\,|\,\|x\|>1\bigr\}\cup\bigl\{x\in X\,|\,\|x\|<1\bigr\}.$$Yes, we can prove that it is an open set, but it is simpler to see that $S$ is the inverse image of the closed set $\{1\}$ with respect to the continuous function $\|\cdot\|$. It follows that $S$ is closed.
A: You can:


*

*show that the norm $N:(X,||\cdot||)\to\mathbb{R}$ mapping $x$ to $||x||$ is a continuous map,

*show that the inverse image of an open set by a continuous map is still open,

*rewrite $S^c$ as $N^{-1}(]-1,1[)\cup N^{-1}(]1,+\infty[)$,

*conclude.
A: Take a convergent sequence on the sphere and shows that it converges there 
Let $(x_{n})$ a convergent sequence in $S$, suppose $x_{n} \to y$
Let's see $y \in S$
$||x_{n}|| = 1$ $ \forall n \in \mathbb{N}$
$||.||$ is a continuos function, then $1 = \lim_{n \to \infty} ||x_{n}|| = ||\lim_{n \to \infty}x_{n}|| = ||y||$ 
Then $y \in S$
So $S$ is a closed set.
A: Even though I think it is much more elegant to use continuity, here is an approach which does not use continuity:
Let first $x_0 \in \{x\in X \mid \Vert x\Vert < 1 \} = S_1^c.$
Take $r = 1-\Vert x_0\Vert > 0.$ Then the open ball
$$B(x_0, r) = \{ y \in X \mid \Vert x_0-y \Vert < r \}$$ is fully contained in $S_1^c$, since for any $y\in B(x_0,r),$ we have
$$ \Vert y\Vert = \Vert x_0 - y\Vert + \Vert x_0 \Vert < r+\Vert x_0 \Vert = 1-\Vert x_0\Vert + \Vert x_0 \Vert = 1. $$
Hence $S_1^c$ is open.
Now let $x_0 \in \{x\in X \mid \Vert x\Vert > 1 \} = S_2^c,$ and take $r = \Vert  x_0 \Vert - 1 > 0.$ Then the open ball $B(x_0,r)$ is fully contained in $S_2^c$, since for any $y\in B(x_0, r)$, the reverse triangle inequality gives
$$ \Vert y \Vert \geq \Vert x_0 \Vert - \Vert y-x_0 \Vert > \Vert x_0 \Vert - r = \Vert x_0 \Vert - (\Vert x_0 \Vert - 1) = 1. $$
Hence $S_2^c$ is open.
Being a union of two open sets, $S^c = S_1^c \cup S_2^c$ is open, so that $S$ is closed.
