Prove that an uncountable subset of a second countable space contains at least one limit point This is a 2 part problem about second countable space.
i)Prove that the set of all isolated points of a second countable space is countable
Definition: $x$ is an isolated point if ${x}$ is open in the topology.
Since the topology is second countable, the set of basis is countable and open sets are defined to be unions of sets in the basis. Thus if x is an isolated point, then x is part of the basis set. Now if there are an uncountable number of isolated points, they are all basis elements, so we would have an uncountable number of basis elements.
(ii) Hence, show that any uncountable subset A of a second countable space contains at least one point which is a limit point of A.
Suppose A does not contain any point that is a limit point. Since the space is second countable, we have a countable basis set $B_i$, $i \in N$ such that their union is the entire space X. Since A is a subset of X, each point of A must be contained in some $B_i$ and each $B_i$ is open. Now I'm not sure if this is correct, but since A is uncountable and we only have countable number of $B_i$, is it the case that one of the $B_i$ must contain an uncountable number of points of A. I think that part is accurate. But how do I go from here to showing that every basis elements that contains points of A all contain another point of A that they all have in common, which would be the limit point of A?
 A: As to (i), you are right: if $X$ is second countable, as witnessed by a countable base $\{B_n\mid n \in \Bbb N\}$ of $X$, than for any isolated point $x$, by the definition of a base there should be a basic set $B_n$ such that $x \in B_n \subseteq \{x\}$ and it follows that $\{x\}=B_n$ is in the base to start with. All singletons of isolated points are members of any base for the space. As there is a countable base, there cannot be more than countably many isolated points.
If then $A$ is an uncountable subset, $A$ itself is a second countable space when given the subspace topology, so $\textrm{Isol}(A)$ (the (relatively) isolated points of $A$) is at most countable by $(i)$ and so $A\setminus \textrm{Isol}(A)$ is uncountable (or $A$ would have been countable, which it is not) and all points in it are by definition limit points of $A$ that (bonus!) are even members of $A$ themselves (there can be many limit points outside as well).
A: (ii) is an entierly trivial consquence of (i).  If there is no limit point then every point is isolated and there are only countable number of them. But then $A$ is countable, a contradiction.
A: If no point of $A$ was a limit point of $A$, then every point of $A$ would be an isolated point, and therefore $A$ would be countable.
