TITLE: Understanding a proof about a set being closed
QUESTION [7 upvotes]: Let $r > 0$ be a positive number, and define $F = \{u \in
 \mathbb{R}^{n} \mid \|u\| \leq r\}$. Prove $F$ is closed in
  $\mathbb{R}^{n}$.

Proof:
We want to show that if a sequence $\{u_{k}\}$ lies in $F$ and $\lim_{k\to\infty} u_{k} = u$, then $u \in F$. Let $\{u_{k}\}$ be in arbitrary sequence in $F$. By the set definition of $F$, it follows that $\|u_{k}\| \leq r$ for each index $k$. Thus,
$$\lim_{k\to\infty} \|u_{k}\| \leq \lim_{k\to\infty} r = r.$$ 
From here, it suffices to show $\lim_{k\to\infty} \|u_{k}\| = \|u\|$. 
The proof goes on and shows that $\lim_{k\to\infty} \|u_{k}\| = \|u\|$

My question:
I don't understand why it suffices to show that the sequence of norms converges to the norm of the limit point? After they prove this fact, they say that $\lim_{k\to\infty} \|u_{k}\| = \|u\| = r$, from which it follows that $u \in F$, hence $F$ is closed. But, I don't get why proving this shows that $u$ is contained in $F$.

REPLY [2 votes]: You can also notice that proving
$$
\lim_{k\to \infty} ||u_k||=||u||
$$
proves that $f(u) =||u||$ is continuous. Now,
$$ 
F=\{ u\in \mathbb R^n : ||u||\le r\} =f^{\leftarrow}([0,r]). 
$$
Recalling that if $f$ is continuous and $C$ is closed then $f^{\leftarrow}(C)$ is closed, the result follows from the fact that $[0,r]$ is closed in $\mathbb R$.