Proof Check: Every compact metric space is separable

Question

Let $X$ be a compact metric space, then it is totally bounded. This means that for $r=1/n$, $n\in \mathbb{N}$ there exist points $x_1,x_2,..,x_{k_{n}}\in X$ such that $$X\subset \bigcup_{i=1}^{k_n}B(x_{i},1/n).$$ Then the set $$E = \bigcup_{n\in \mathbb{N}}\{x_1,x_2,.,,,x_{k_n}\}\subset X$$ and is countable. I want to show that $E$ is dense in $X.$ For that let $x\in X.$ If $x\in E$ then we are done. Otherwise we want to show that $x$ is a limit point of $E.$ So let $B(x,r)$ be a ball centered at $x$ of radius $r>0.$ Then there exists an integer $m$ such that $B(x,1/m)\subset B(x,r).$ Now when $n=m$ every point in $X$ is at most $1/m$ units away from at least one point in the set $\{x_1,x_2,...,x_{k_m}\}$ so at least one of the points $x_j\in B(x,1/m)\subset B(x,r)$ different from $x.$ Thus $x$ is a limit point of $E.$ So $E$ is dense in $X.$ Is this proof correct?

Answer 1

It is better to write $E=\{x_1^n,...,x_{k_n}^n\}$. You have the idea. But you can write the converging sequence like so: for every $x$, and every integer $n>0$, $x\in B(x_j^n,1/n)$, this implies that $lim_nx_j^n=x$.

Answer 2

Alternatively, if $U$ is any nonempty open set in $X$ and $x\in U$, then you can take some $n$ with $B(x,1/n)\subseteq U$. Then take some $i\in\{1,\dots,k_n\}$ for which $x\in B(x_i,1/n)$. It follows that $x_i\in B(x,1/n)\subseteq U$, so your set $E$ intersects every nonempty open set in $X$.

Also as the other responder pointed out, the notation is slightly ambiguous. For each $n$ you can label the $x_i$ as $x_{n,i}$ for $i=1,\dots,k_n$. Then the result of our argument above would be that $x_{n,i}\in U$ for some $i\in\{1,\dots,k_n\}$.