Deduce that the unit ball $\{ \textbf{x} \in \mathbb{R^n} : \| \textbf{x} \| \leq 1 \}$ is closed 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?
 A: For (ii): You said that $x = \lim\limits_{k \to \infty} x_{k}$ implies that for all $k$, $\|x_{k}\| \leq 1$. While the statement $\|x_{k}\| \leq 1$ is true, it is because each $x_{k}$ is already a member of the unit ball. For the implication to be true, you would have to already know that $\|x\| \leq 1$, but this is what you were trying to prove. 
Here is an alternative proof of (ii): 
Let $x$ be a limit point of $U$. By definition, for all $\epsilon > 0$ there exists some $y \in U$ such that $\|x - y\| < \epsilon$. So there exists a sequence $(x_{k})$ of points of $U$ such that for each $k$, $\|x_{k} - x\| < 1/k$. Thus $\lim\limits_{k \to \infty} x_{k} = x$, so 
$\|x\|= \lim\limits_{k \to \infty} \|x_{k}\|$ by part (i).
Since each $x_{k} \in U$, $\|x_{k}\| \leq 1$. From the reverse triangle inequality it follows that 
$$\|x\| - \|x_{k}\| \leq \| x - x_k \| < 1/k$$
Since $\|x\| - 1 \leq \|x\| - \|x_{k}\|$, we can combine these inequalities to obtain that for all $k$, $$\|x\| < 1 + 1/k$$
Thus $\|x\| \leq 1$, so $U$ contains all limits points and is closed.
A: 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
$$
\lVert \textbf{x} \rVert\leq 1.
$$
