Let $X$ be a separable normed space and $ A \subseteq X$ be nonempty. Does there exists a countable subset $A’$ of $A$ such that $A’$ is dense on $A$? If not, please provide a counterexample. What about the case on which $X$ is finite dimensional?
It seems simple but I have not being able to came up with a general construction, even in finite dimensions. Initially I thought of taking the smallest subspace containing $A$ and defining $A’: = S \cap ric(A)$, where $S$ is a countable dense subset of $X$ and $ric(A)$ denotes the relative interior of $A$, but this doesn't work.
Best Answer
Any separable metrisable space is second-countable. Second-countability is hereditary and implies separability. So every subset of a separable metrisable space is separable.
Suppose $S$ is a countable dense subset of $X$. For each $s \in S$ and $n \in \mathbb N$ choose, if possible, $a_{sn} \in A$ with $||s - a_{sn}|| < 1/n$. The set of these $a_{sn}$s is a countable dense subset of $A$.