mjhjmtu

Let $r > 0$ be a positive number, and define $F = u$. Prove $F$ is closed in
$mathbbR^n$.

Proof:

We want to show that if a sequence $u_k$ lies in $F$ and $lim_ktoinfty u_k = u$, 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,

$$lim_ktoinfty |u_k| leq lim_ktoinfty r = r.$$

From here, it suffices to show $lim_ktoinfty |u_k| = |u|$.

The proof goes on and shows that $lim_ktoinfty |u_k| = |u|$

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 $lim_ktoinfty |u_k| = |u| = r$, 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 $F$.

I think you are a bit confused so I'll try to make up the argument with more details. First, showing that a set is closed is equivalent to showing that every converging sequence has a limit point IN the set. Here the set we are concerned with is $F=leq r$. Thus, if we pick a sequence $(u_k)_ksubset F$ such that $u_kto u$, it suffices to show that $uin F$ to prove that $F$ is closed. What does it mean to be an element of $F$ ? It means simply that $||u||leq r$ ! So we have to show that $||u||leq r$ :) That's why it is sufficient to show that the norm of the $u_k$'s goes to the norm of $u$ because $||u_k||leq r$ for all $k$ and by taking the limit inside the norm because the norm is continuous, we get $||u||leq r$ which is the condition an element of $mathbbR^n$ has to satisfy to be in $F$.
If you want more details, let me know !

The proof provided by Malik is correct. I'm giving a separate answer because there's another approach that often works in general and is often easier. A set is closed if its complement is open. So assume you have a point $x notin F$, so that $||x|| gt r$. Can you prove that there's necessarily an open ball around $r$ that does not intersect $F$?

Hint: Let $||x|| = r+a, a gt 0$. Let $epsilon = a/2$ and consider $B(x, epsilon)$ using the triangle inequality.

You can also notice that proving
$$
lim_kto infty ||u_k||=||u||
$$
proves that $f(u) =||u||$ is continuous. Now,
$$
F=le r =f^leftarrow([0,r]).
$$
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 $mathbb R$.