Deduce that the unit ball $\{ \textbf{x} \in \mathbb{R^n} : \| \textbf{x} \| \leq 1 \}$ is closed

Question

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?

Answer 1

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.

Answer 2

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 $$ \lVert \textbf{x} \rVert\leq 1. $$