Completeness of a metric space $(X,d)$ w.r.t the sup metric

Question

$\mathbf{Question}$: Let $X=\{(x_1,x_2,...): x_i \in \mathbb{R}$ and only finitely many $x_i$'s are non-zero $\}$ and $d:X \times X \to \mathbb{R}$ be a metric on $X$ defined by $d(x,y)=\sup_{i \in \mathbb{N}} |x_i-y_i|$ for $x=(x_1,x_2,...), y=(y_1,y_2,...)$ in $X$.

$1.$ Is $(X,d)$ a complete metric space?

$2.$ Is the set $\{x \in X: d(\underline{0},x) \leq 1\}$ compact? [$\underline0$ is the zero element of $X$].

$\mathbf{Attempt}$:

Consider the sequence $\{x_n\}$ defined by $x_n=(1,\frac{1}{2},..,\frac{1}{n},0,0,...)$ , where $\frac{1}{n}$ is at the $n^{\text{th}}$ place.

It is indeed a Cauchy Sequence in $(X,d)$, since $\forall \epsilon>0$ $\exists $ a $K \in \mathbb{N}$ such that $d(x_n,x_m)=|\frac{1}{m+1}|<\epsilon$ , $\forall n> m>K$ (without loss of generality, and for $m=n$, ineq is trivial), $K=\lceil \frac{1}{\epsilon}\rceil+1$.

Now, $x_n\in l^{\infty}$, $\forall n \in \mathbb{N}$ and the sequence being Cauchy, it converges to some point, say $\lambda=(l_1,l_2,...)$. However, $\lambda \notin X$, as $\forall k \in \mathbb{N}$, $l_k=\frac{1}{k} \neq 0$.

Therefore, $(X,d)$ isn't complete.

We observe that $\forall n \in \mathbb{N}$, $d(\underline{0},x_n)\leq 1$, yet $\{x_n\}$ converges to $\lambda \notin \{x \in X: d(\underline0,x) \leq 1\}$, showing that it is not closed. Consequently, it isn't compact.

Is my proof okay?

Kindly $\mathbf{VERIFY}$.

Answer 1

It is correct. But you used a fact that you did not need to use. Although $\ell^\infty$ is indeed complete, it is easy to see that your sequence converges to $\lambda$ and so you do not need to assume that it converges.

On the other hand, note that the fact that set $\{x\in X\mid d(0,x)\leqslant1\}$ (which is a closed subset of $X$) is not a closed subset of $\ell^\infty$ proves both statements that you want to prove: if $X$ was complete, then $\{x\in X\mid d(0,x)\leqslant1\}$ would be complete too and therefore it would be a closed subset of any larger metric space, and if $\{x\in X\mid d(0,x)\leqslant1\}$ was compact then, again, it would be a closed subset of any larger metric space.

Answer 2

Instead of saying that there is a limit in a larger space, but that limit is not $\in X$, is would prefer to argue that every $y\in X$ fails to be a limit of your sequence. Indeed, if $y_i=0$ for all $i\ge N$, then $d(x_n,y)\ge \frac1n$ for all $n>N$, so $x_n\not\to y$.