Consider the space of absolutely convergent functions $$X:=\left\{(x_i)_{i=0}^{\infty}:\sum_{i=0}^{\infty}|x_i|<\infty\right\}$$ and define a metric $d$ on $X$ by $$d((x_i)_{i=0}^{\infty},(y_i)_{i=0}^{\infty}):=\sum_{i=0}^{\infty}|x_i-y_i|.$$
Now for each $n\in \mathbb{N}$, define the sequence $e^{(n)}:=(e^{(n)}_i)_{i=0}^{\infty}$ in $X$ such that $e^{(n)}_{i}:=1,$ if $n=i$ and $e^{(n)}_{i}:=0,$ otherwise.
Let $Y:=\{e^{(n)}:n\in \mathbb{N}\}\subset X$. I would like to show that $Y$ is a closed and bounded subset of $X$, but it is not compact.
I have managed to show that $Y$ is bounded, but I am kind of stumped on how to show that it is closed but not compact.
Any help/hint will be useful. Thanks in advance.
What is the distance $d(e^{(n)},e^{(m)})$? Recall that a space is compact if and only if every open cover contains a finite subcover. What can you say about the open balls $B_{\epsilon}(e^{(n)})$ for $\epsilon>0$?
There are several ways to show $Y$ is closed. One possibility would be to show that, for any $x$ not in $Y$, there is an $\epsilon_{x}>0$ such that $B_{\epsilon_{x}}(x)$ is disjoint from $Y$. You can then note that $\bigcup_{x\notin Y}B_{\epsilon_{x}}(x)=X\setminus Y$ is open.