A closed countable set with Cantor-Bendixson rank of $\omega +1$

Question

I'm looking to find a closed countable set that has a Cantor-Bendixson Rank of $\omega +1$.

I know that $\{0\}\cup\{\frac{1}{x+1}|x\in\omega\}$ has a Cantor-Bendixson Rank of $2$ because we take out the isolated points and are left with $\{0\}$.

I also know how to get sets with a C-B rank of $n,n\in\omega$ by iteratively doing this process.

I know that the solution $X$ will have $X^{(\omega +1)}=\emptyset$ and $X^{(\omega )}$ is a set of isolated points.

Also $$X^{(\omega )}=\bigcap_{y\in\omega }X^{(y)}$$

But I'm unsure how to proceed.

Answer 1

Start with $\{0\}\cup \{\frac{1}{n+1}\mid n\in \omega\}$. Now for each $n\in \omega$, add points in the interval $(\frac{1}{n+1},\frac{1}{n})$ limiting to $\frac{1}{n+1}$, to ensure that $\frac{1}{n+1}$ is isolated in $X^{(n)}$, but not in $X^{(n-1)}$. (In other words, you want to make $X\cap [\frac{1}{n+1},\frac{1}{n})$ into a space of Cantor-Bendixson rank $n+1$, which you say you know how to do in your question.)

Then $\bigcap_{n\in \omega} X^{(n)} = \{0\}$, so $X$ has Cantor-Bendixson rank $\omega+1$.