Is the cardinality of the set of all isolated points in a second countable metric space at-most $\aleph_0$?

Question

Is the cardinality of the set of all isolated points in a second countable metric space at-most $\aleph_0$ ?

Answer 1

A second-countable metric space $X$ is hereditarily Lindelof, that is, every subspace is Lindelof. I would prove a stronger result, namely if $Z\subseteq X$ has the property that every $z\in Z$ has a neighborhood $U_z$ such that $U_z\cap Z$ is countable, then $Z$ must be countable. Indeed, for each $z\in Z$ fix such a neighborhood $U_z$ and let $U=\cup\{U_z:z\in Z\}$. Then the family $\mathcal U=\{U_z:z\in Z\}$ is an open cover $U$ and hence must have a countable subcover, say $\{U_z:z\in Z_0\}$ for some countable $Z_0\subseteq Z$. Then $Z=\cup_{z\in Z_0} U_z\cap Z$, hence $Z$ is countable.

As a corollary, in a second-countable metric space $X$, if $Y$ is an uncountable subset, then we could throw out at most countably many points of $Y$, say these points form a set $Z$, so that every neighborhood of every $y\in Y\setminus Z$ intersects $Y\setminus Z$ in uncountably many points.

Answer 2

"Second countable" means that your space has a countable base $\mathcal B$ for its topology. If a point $x$ is isolated, then $\{x\}$ is open, hence a union of sets from $\mathcal B$, and hence an element of $\mathcal B$.