TITLE: Deduce that the unit ball $\{ \textbf{x} \in \mathbb{R^n} : \| \textbf{x} \| \leq 1 \}$ is closed
QUESTION [2 upvotes]: Show that, if $(\textbf{x}_{k})$ is a sequence in $\mathbb{R^n}$ with $\textbf{x}_{k} \rightarrow \textbf{x}$, then $\lVert \textbf{x}_k \rVert \rightarrow \lVert \textbf{x} \rVert$.
Deduce that the unit ball $U =\{ \textbf{x} \in \mathbb{R^n} : \lVert \textbf{x} \rVert \leq 1 \}$ is closed
Attempted Solution:
i) if $\textbf{x}_{k} \rightarrow \textbf{x}$ then by definition, for all $\epsilon > 0$ there exists $K \in \mathbb{N}$ such that when $k \geq K$ then $$ \lVert \textbf{x}_k - \textbf{x} \rVert \leq \epsilon $$
By the reverse triangle inequality:
$\lVert \textbf{x}_k \rVert - \lVert \textbf{x} \rVert 
 \leq \lVert \textbf{x}_k - \textbf{x} \rVert \leq \epsilon $
This takes care of the first part. But more importantly I was having trouble with part ii)
ii)
Attempt: 
Let $\textbf{x}$ be a limit point of  $U$. 
$\Rightarrow$ there exists a sequence $(\textbf{x}_{k})_{k = 1}^{\infty}$ with $\textbf{x}_k \in U$ such that $\textbf{x} = \lim_{k \rightarrow \infty} \textbf{x}_k$
Since $\textbf{x}_{k} \rightarrow \textbf{x}$ by part i):
$\lVert \textbf{x}_k \rVert \rightarrow \lVert \textbf{x} \rVert $
this means $\lVert \textbf{x}_k \rVert \leq 1$
Therefore $U$ contains all of its limit points, so $U$ is closed.
Is my reasoning for these solutions be correct?

REPLY [1 votes]: I think you didn't complete the proof for the i) proposition.
Besides the inequality
$$
\lVert \textbf{x}_k \rVert - \lVert \textbf{x} \rVert 
 \leq \lVert \textbf{x}_k - \textbf{x} \rVert \leq \epsilon 
$$
you also have
$$
\lVert \textbf{x} \rVert-\lVert \textbf{x}_k \rVert   
 \leq \lVert \textbf{x}_k - \textbf{x} \rVert \leq \epsilon 
$$
Combining both, you get
$$
\lVert \mathbf{x}\rVert-\epsilon\leq\lVert \textbf{x}_k \rVert\leq \lVert \textbf{x} \rVert+\epsilon
$$
ii) From the second form of the reversed triangle inequality, you obtain
$$
\lVert \textbf{x} \rVert\leq \lVert \textbf{x}_k \rVert+\epsilon\leq 1+\epsilon,
$$
since $\mathbf x_k\in U$.
As this is valid for all $\epsilon>0$, you obtain
$$
\lVert \textbf{x} \rVert\leq 1.
$$
{"set_name": "stack_exchange", "score": 2, "question_id": 3002758}