# If $X$ is an unbounded subset of $\omega_{1}$, then the set of all countable limit points of $X$ is a closed unbounded set.

I have been reading the book "Introduction to Set Theory" by Jech and Hrbacek and have come to an exercise I'm having difficulty with in the chapter on filters and ultrafilters. The exercise comes from section 3 of the chapter on closed, unbounded, and stationary sets.

The exercise goes :

If $$X$$ is an unbounded subset of $$\omega_{1}$$, then the set of all countable limit points of $$X$$ is a closed unbounded set.

Before the exercise is the text :

If $$X$$ is a set of ordinals, then $$\alpha$$ is a limit point of $$X$$ if for every $$\gamma < \alpha$$ there is $$\beta \in X$$ such that $$\gamma < \beta < \alpha$$. A countable $$\alpha$$ is a limit point of $$X$$ if and only if there exists a sequence $$\alpha_{0} < \alpha_{1} < \dots$$ in $$X$$ such that $$\sup\left( \{ \alpha_{n} \; \mid \; n \in \omega \} \right) = \alpha$$. Every closed unbounded $$C \subseteq \omega_{1}$$ contains all its countable limit points.

$$\newcommand{\Suc}{\text{Suc}} \newcommand{\Lim}{\text{Lim}} \newcommand{\Ord}{\text{Ord}}$$

Here I use the notation $$\Suc$$ for the successor ordinals, $$\Lim$$ for the limit ordinals, and $$\Ord$$ for the class of all ordinals.

Here is my attempt at a solution thus far :

Let : $$$$A = \{ \alpha \in \Ord \; \mid \; \alpha \text{ is a limit point of } X \}$$$$ First show : $$$$\alpha \in A \Rightarrow \alpha \not \in \Suc$$$$ Let's suppose : $$$$\alpha \in A \text{ and } \alpha \in \Suc$$$$ Let : $$$$\alpha = \gamma + 1$$$$ So : $$$$\gamma < \alpha \text{ and } \not \exists \beta \in \Ord \text{ s.t. } \gamma < \beta < \alpha$$$$ and since $$X \subseteq \Ord$$ : $$$$\gamma < \alpha \text{ and } \not \exists \beta \in X \text{ s.t. } \gamma < \beta < \alpha$$$$ So $$\alpha$$ cannot be a limit point of $$X$$ when $$\alpha \in \Suc$$. $$\checkmark$$

So : $$$$\alpha \in A \Rightarrow \alpha \not \in \Suc \; \checkmark$$$$ and : $$$$A = \{ \alpha \in \Lim \; \mid \; \alpha \text{ is a limit point of } X \}$$$$ We know since $$X$$ is unbounded : $$$$\sup(X) = \omega_{1}$$$$ We know : $$$$\alpha \in X \Rightarrow \exists \beta \in A \text{ s.t. } \alpha \leq \beta$$$$ Since every $$\beta \in A$$ is the supremum of some increasing sequence of length $$\omega$$ in $$X$$. So : $$$$\sup(X) \leq \sup(A)$$$$ and since : $$$$\alpha \in A \Rightarrow \alpha < \omega_{1}$$$$ We know : $$$$\sup(A) \leq \omega_{1}$$$$ So : $$$$\omega_{1} = \sup(X) \leq \sup(A) \leq \omega_{1} \Rightarrow \sup(A) = \omega_{1} \; \checkmark$$$$ and $$A$$ is unbounded. $$\checkmark$$

I am not sure how to show that $$A$$ is also closed. Can someone help with this ?

• I’m missing something here. Isn’t the set of all countable successor ordinals, augmented by at least $1$ but no more than countably many countable limit ordinals, a counterexample? Don’t we also need $X$ closed? Commented Dec 11, 2021 at 23:56
• @RobertShore Not sure what you mean by "augmented" here, but the statements as it is now is correct (note that the set of limit points of $X$ can be even disjoint from $X$)
– ℋolo
Commented Dec 12, 2021 at 0:03
• @ℋolo Ah. I was reading "limit points of $X$" to require that they be within $X$. Since the union of a set of ordinals with its limit points is closed, this is asking for a proof that the set of limit ordinals in a closed unbounded set is itself closed and unbounded. Commented Dec 12, 2021 at 6:07

Closed means that any limit of elements of $$A$$, are also in $$A$$, so let's take increasing a sequence $$(α_i ∈ A)_{i∈ω}$$, and we will show that $$\lim α_i=α∈ A$$ (I assume monotocity, so $$\lim$$ here is the same as $$\sup$$).

We know that each $$α_i$$ is a limit of some increasing sequence $$(β_i^j ∈ X)_{j\in\omega}$$, define $$f:ω→ω$$ to be $$f(i)=\min(j\in\omega\mid β_{i+1}^j>α_i)$$ (note, it is well defined because $$α_i=\lim_j β_i^j<\lim_j β_{i+1}^j=α_{i+1}$$, also note that $$f$$ is not necessarily increasing).

I claim that $$\lim_i β_{i+1}^{f(i)}=α$$, indeed $$β_{i+1}^{f(i)}<α_{i+1}<α$$ for all $$i$$, so $$\lim β_{i+1}^{f(i)} ≤α$$, but if $$x<α$$, there exists $$i$$ such that $$x<α_i<α$$, and so $$x<α_i<β_{i+1}^{f(i)}<α$$, hence $$\lim_i β_{i+1}^{f(i)}=α$$.

So, given any limit point $$α$$ of $$A$$, we have a (countable) sequence from $$X$$, whose limit is $$α$$, hence $$α∈A$$

• To me it looks like $(\beta_{j}^{i} \in X)_{j \in \omega}$ should be $(\beta_{i}^{j} \in X)_{j \in \omega}$. Is this correct ? Commented Dec 12, 2021 at 17:57
• @scipio yes, thanks I fixed it
– ℋolo
Commented Dec 13, 2021 at 14:55

If $$b\in A'$$ then there is a sequence $$(a_i)\subseteq A$$ such that $$\sup (a_i)=b.$$ On the other hand, since $$a_i\in X'$$ for each integer $$i$$, there is a sequence $$(b_i)\subset X$$ such that $$a_1\le b_2\le a_2\le b_3\le a_3\le \cdots$$.

If $$\sup (b_i)=\beta then there is an integer $$i$$ such that $$\beta< a_i\le b$$ and therefore $$\beta< a_i\le b_{i+1}\le b.$$ Thus, $$\beta which is impossible. We conclude that $$\sup (b_i)=\beta=b$$ and so $$b\in X'.$$

Remark: We used the fact that $$b$$ is $$\textit{actually in}\ \omega_1.$$ To prove this,

Without loss of generality, we may assume that for all integers $$i$$, some $$j > i$$ satisfies $$a_i < a_j.$$ Set $$Y_i=\{\beta\le a_i\}.$$ Then, $$Y=\bigcup_{i\in \omega} Y_i$$ is countable so $$\omega_1\setminus Y\neq \emptyset.$$ Let $$b$$ be least in $$\omega_1\setminus Y.$$ Then, for each integer $$i,\ a_i\le b$$ so $$b$$ is an upper bound for $$(a_i)$$; and if $$\delta then $$\delta\in Y.$$ It follows that there are integers $$i,j$$ and $$\beta\in Y_i$$ such that $$\delta=\beta\le a_i That is, $$\delta$$ is not an upper bound for $$(a_i).$$ Hence $$b=\sup (a_n).$$