Jtdylktuy

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 $lim_{ktoinfty} u_{k} = u$, 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 $F={uin mathbb{R}^n : ||u||leq r}$. 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 $mathbb{R}^n$ 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,
$$
F={ uin mathbb R^n : ||u||le r} =f^{leftarrow}([0,r]).
$$
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 $mathbb R$.