Altering separable space's definition

Question

A separable space has by definition a countable sub-space that is dense.

What if we replace "dense" by "its completion is dense"

What about "its completion is the whole space"? Do such spaces have names (if they are different)?

I guess my question only applies to metric spaces with the associated topology

Answer 1

If there is a countable subset whose completion is the given metric space then the space is separable. But the converse is not true because a metric space can be separable without being complete.

Answer 2

(1).If Y is a subspace of a separable metric space X then Y is separable. So if X is the metric completion of Y, and X is separable, then Y is separable.

(2).If Y is a dense separable subspace of X then X is separable. So if Y is a separable metric space then the metric completion of Y is separable.

For (1) let $d$ be a metric for $X$ and let $D$ be a countable dense subset of $X.$ Let $E\subset D\times \Bbb Q^+,$ where $e=(\delta,q)\in E\iff B_d(\delta,q)\ne \phi.$ Let $f:E\to Y$ be a function with $f(\delta,q)\in Y\cap B_d(\delta,q).$

Now $F=\{f(e):e\in E\}$ is a countable subset of $Y.$ To show that $F$ is dense in $Y:$ If $y\in U\subset Y,$ where $U$ is open in $Y,$ then take $r\in \Bbb R^+$ such that $Y\cap B_d(y,r)\subset U.$

Now take $q\in \Bbb Q^+$ with $q<r/3.$ Since $D$ is dense in $X$ there exists $\delta \in D \cap B_d(y,q),$ so $y\in B_d(\delta,q)\cap Y,$ so $(\delta, q)\in E.$ And by the triangle inequality (since $3q<r$ and $d(y,\delta)<q $ ) we have $B_d(\delta,q))\subset B_d(y,r).$ So $F\cap U$ is not empty because $$f(\delta,q)\in F\cap B_d(\delta,q)\subset F\cap B_d(y,r)=$$ $$=(F\cap Y)\cap B_d(y,r)=F\cap (Y\cap B_d(y,r))\subset F\cap U.$$

For (2), if $G$ is countable dense subset of $Y$ and Y is dense in $X$ then $$ X=Cl_X(Y)=Cl_X(\,Cl_Y(G)\,)=Cl_X (\,Y\cap Cl_X(G)\,)\subset Cl_X(\,Cl_X(G)\,)=$$ $$=Cl_X(G)\subset X.$$ So $X=Cl_X(G).$