Is this a typo about the derived set in textbook Introduction to Set Theory by Hrbacek and Jech?

Question

In my textbook, $A'$ is the set of all limit points of $A \subseteq \Bbb R$.

I think the below statement is possibly not correct.

For each $\alpha$, the set $F^{(\alpha)} - F^{(\alpha+1)}$ is the set of ALL isolated points of $F^{(\alpha)}$.

We know that $F^{(\alpha)} \subseteq F$ and $F^{(\alpha)}$ is a closed set for all ordinal $\alpha$. Thus for all $x \notin F$, $x \notin F^{(\alpha)}$ and thus $x$ is an isolated point of $F^{(\alpha)}$. As such, $F^{(\alpha)} - F^{(\alpha+1)}$ is the set of isolated points of $F^{(\alpha)}$, but not ALL.

Please confirm if my observation is correct or not! Thank you for your help!

Answer 1

Looks right.

It $x\in F^{\alpha}$ is not a limit point, it's an isolated point (of $F^{\alpha}$). But $F^{\alpha+1}$ is by definition the derived set of $F^{\alpha}$. Thus $F^{\alpha}\setminus F^{\alpha+1}$ is $F^{\alpha}$ minus its limit points. What's left is the isolated points.

As pointed out in the comments, points outside a set are not considered isolated points (of the set).

Answer 2

For any set $A$, the set $A\setminus A'$ is the set of isolated points of $A$:

$x$ is in $A-A'$ if $x \in A$ and $x \notin A'$ so there exists an open neighbourhood $O$ of $x$ such that $O \cap A \subseteq \{x\}$ and as we already know $x \in A$, we know that $O \cap A = \{x\}$, i.e. $x$ is an isolated point of $A$. The reverse is obvious: $x$ is an isolated point of $A$ iff $x \in A$ and there is an open set $O$ of $X$ such that $O \cap A= \{x\}$ (i.e. $\{x\}$ is relatively open in $A$), which shows $x \notin A'$ too. So $x \in A-A'$.

The statement in the book is just the special case $A=F^{\alpha)}$ because $A'= F^{(\alpha+1)}$ by the definition.

So no typo, it's completely correct.