Example of H-set

Question

A subset $Z$ of a topological space $X$ is said to be an $\textbf{H-set}$ if there exists a transfinite decreasing sequence $\{ F_{\sigma}:\alpha < \kappa \}$ of closed subsets of $X$ such that $Z=\bigcup \{ F_{\alpha} \backslash F_{\sigma +1} :\alpha < \kappa, \alpha \text{ even ordinal} \}$.

I try to understand the definition by constructing an example.

Let $X=\mathbb{R}$ and $Z=(0,1)$. I don't know what sequence to define here. I define $F_\alpha=[\frac{1}{\alpha},1-\frac{1}{\alpha}]$ for $\alpha \geq 2$. But clearly this does not work.

Answer 1

Note that $\kappa$ is not required to be infinite. You can take $\kappa=2$, $F_0=\Bbb R$, and $F_1=\Bbb R\setminus(0,1)$, so that $(0,1)=F_0\setminus F_1$. Alternatively, you can make $\kappa=\omega$ by letting $F$ be any closed subset of $F_1$ and setting $F_n=F$ for $n\ge 2$.

The following equivalence may make the idea a little easier to think about. First we have the

Definition. If $X$ is a topological space, a set $H\subseteq X$ is an $H$-set (or a resolvable set) if there are an even ordinal number $\kappa$ and closed sets $F_\alpha$ for $\alpha<\kappa$ such that $$H=\bigcup\{F_\alpha\setminus F_{\alpha+1}:\alpha<\kappa\text{ and }\alpha\text{ is even}\}\;.$$

Then it’s not hard to show that

$H\subseteq X$ is an $H$-set iff there are a sequence $\langle C_\alpha:\alpha\le\lambda\rangle$ of closed sets such that

• $C_0=X$,
• $C_\alpha\supseteq C_{\alpha+1}$ for all $\alpha<\lambda$,
• $C_\eta=\bigcap_{\alpha<\eta}C_\alpha$ is $\eta$ is a limit ordinal, and
• $C_\kappa=\varnothing$,

and an $I\subseteq\lambda$ such that $H=\bigcup_{\alpha\in I}(C_\alpha\setminus C_{\alpha+1})$.

Suppose first that $H$ is an $H$-set, and let $\{F_\alpha:\alpha<\kappa\}$ be as in the definition. By using the trick described in the first paragraph, we may assume that $\kappa$ is a limit ordinal. Let $C_0=X$. For $\alpha<\kappa$ let $C_{\alpha+1}=F_\alpha$, and for each limit ordinal $\eta\le\kappa$ let $C_\eta=\bigcap_{\alpha<\eta}C_\alpha$. Finally, let $\lambda=\kappa+1$ and $C_\lambda=\varnothing$. Then $\langle C_\alpha:\alpha\le\lambda\rangle$ and the set $I=\{\alpha+1:\alpha<\kappa\text{ and }\alpha\text{ is even}\}$ have the required properties.

Now suppose that we have $\langle C_\alpha:\alpha\le\lambda\rangle$ and $I\subseteq\lambda$ as described. Let $I=\{\iota_\xi:\xi<\beta\}$, where $\iota_\xi<\iota_\eta$ whenever $\xi<\eta<\beta$. Let $F_0=C_{\iota_0}$ and $F_1=C_{\iota_0+1}$, so that $F_0\setminus F_1=C_{\iota_0}\setminus C_{\iota_0+1}$. Let $F_2=C_{\iota_1}$ and $F_3=C_{\iota_1+1}$, so that $F_2\setminus F_3=C_{\iota_1}\setminus C_{\iota_1+1}$. In general, if $\xi<\beta$, and $\xi=\eta+n$, where $\eta$ is $0$ or a limit ordinal, and $n\in\omega$, let $F_{\eta+2n}=C_{\iota_\xi}$ and $F_{\eta+2n+1}=C_{\iota_\xi+1}$. Then $F_{\eta+2n}\setminus F_{\eta+2n+1}=C_{\iota_\xi}\setminus C_{\iota_\xi+1}$, and the union of the sets $F_\alpha\setminus F_{\alpha+1}$ over even $\alpha$ is equal to $H$.

Now you can avoid worrying about odd versus even indices.