Numerability on subset of $\mathbb{R}$ with induced topology A friend of mine has this statement, that I cannot establish wether it is true or false. Let us consider $\mathbb{R}$ with the euclidean topology. Let us consider $A \subset \mathbb{R}$ such as the topology induced on it is the discrete topology, so $A$ is numerable. Actually he states that the statement holds whenever $A$ is closed, but I ignore his proof. I tried to demonstrate, recreating something like the Cantor set, that it is false proved that there exists a non separable set $A$ with that induced topology, but I cannot prove the existence of such $A$. With discrete topology I mean that for every $x \in A$ there exists an open neighborhood of $x$, that I'll call $I(x)$ such that $I(x) \cap A = {x} $. Any suggestions?

For each $p\in A$ there is a $U_p$ such that $A \cap U_p = \\{p\\}$; the collection $\\{U_p\\}$ is uncountable. In each $U_p$ we can find an interval $V_p$ with rational endpoints containing $p$. Since also $A \cap V_p = \\{p\\}$, the $V_p$ are distinct and the collection $\\{V_p\\}$ is uncountable. But the collection of intervals $(q,r)$ with rational endpoints is in bijection with ordered pairs of rational numbers, of which there are only countably many.