Second countable scattered spaces are countable Let $X$ be a second countable space and $X$ is scattered. > **A scattered space** is a space for which every not empty subset has an isolated point. **How to show $X$ is countable?** Thanks ahead.

Suppose that $X$ is uncountable. Let $\mathscr{B}$ be a countable base for $X$. Let $\mathscr{B}_0=\\{B\in\mathscr{B}:|B|\le\omega\\}$, let $A=\bigcup\mathscr{B}_0$, and let $Y=X\setminus A$. Then $|A|\le\omega$, so $Y$ is uncountable. Let $y\in Y$ be arbitrary. Suppose that $B\in\mathscr{B}$ and $y\in B$; then $B\
subseteq A$, so $B\
otin\mathscr{B}_0$, and therefore $|B|>\omega$. Finally, $|A|\le\omega$, so $|B\cap Y|=|B\setminus A|>\omega$. Thus, not only is $y$ not an isolated point of $Y$, but in fact every open nbhd of $y$ contains uncountably many points of $Y$.