Boundary of unit ball is closed and bounded

Question

Consider the space

$$M= \{\textbf{z} \in \mathbb{R}^N : \|\textbf{z}\|_1=1 \}.$$

Show that $M$ is closed and bounded.

Let $(\textbf{z}_n)$ be a sequence of vectors which converge to some $\textbf{z} \in \mathbb{R}^n$. I must show $\|\textbf{z}\|_1=1$.

Indeed, since $\|\cdot\|_1$ is continuous, $\|\textbf{z}_n\|_1\rightarrow \|\textbf{z}\|_1$.

But $\textbf{z}_n$ is in $M$, hence it is equal to $1$, a constant. Therefore, $\|\textbf{z}\|_1=1$.

Boundedness: $\|x-y\|_1\leq \|x\|_1+\|y\|_1=2$. Since $x,y$ are arbitrary, $\operatorname{diam}M<\infty$.

Is this reasonable?

Answer 1

Or you could say that it's the inverse image of $1$ under the continuous norm function, hence closed. And clearly it's bounded, since it's contained in a ball of radius $2$.

Answer 2

Your proof is correct and sounds very reasonable.