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Let $S_\Omega$ be an uncountable well-ordered set every initial segment of which is countable. Consider $S_{\Omega}$ with the order topology (i.e. its well-order). Let $C \subset S_{\Omega}$. Show that the following are equivalent.

(1) $C$ is closed.

(2) For each $A \subset C$ not empty and at most countable, $sup A \in C$

All I know about $S_{\Omega}$ is that its initial segments are all at most coutable.

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    Try showing that $\sup A$ is a limit point of $A$, if its not $\max A$. This holds in all ordered spaces – 2017-02-16

2 Answers 2

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Suppose that $C$ is not closed. This means there is some $\alpha \in S_\Omega$ that is in $\overline{C}\setminus C$. For any $y < \alpha$ the set $O(y):=(y, \alpha+1)$ is basic open in the order topology on $S_\Omega$. It contains $\alpha$ so these all intersect $C$, as $\alpha \in \overline{C}$. Also as we live in $S_\Omega$, there are only countably many ordinals smaller than $\alpha$. So it has countable cofinality, and we can find a sequence $\beta_n < \alpha$ such that $\beta_n$ increases to $\alpha$. Then pick $a_n \in O(\beta_n) \cap C$,and set $A = \{a_n: n \in \mathbb{N}\}$. Then $A \subset C$ is countable but $\sup A = \alpha \notin C$, so the condition fails. So the countable sup-condition implies closedness.

The reverse direction is true in all ordered spaces. Closed sets are closed under suprema of subsets.

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(1) $\Rightarrow$ (2): First, if $C$ is closed, and $A \subset C$ is bounded above, then $\sup(A) \in C$. So you just need to show that $A$ is bounded above if $A$ is countable. To do so, we see:

$$L := \{x \in S_{\Omega} \mid \exists y \in A: x \leq y\} = \bigcup_{y \in A}[0,y]$$

(where $[0,y]$ denotes the initial segment determined by $y$) is a countable union of countable sets, and therefore countable. So $S_{\Omega} \backslash L$ is nonempty, and every element in $S_{\Omega} \backslash L$ is an upper bound of $A$.

(2) $\Rightarrow$ (1): We can show the complement of $C$ is open. Pick $y \in S_{\Omega} \backslash C$, and let $A = C \cap [0,y]$. Then $A$ is a countable subset of $C$, so $x := \text{sup}(A) \in C$. If we let $s(y)$ denote the successor of $y$, we see that the open interval $(x,s(y))$ is disjoint from $C$.

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    I didn't understand some parts. Why "First, if $C$ is closed, and $A\subset C$ is bounded above, then $sup(A)\in C$. " Everytime the supremum is a accumulation point? And, also, the second stetement proved that the complement of C is open. – 2017-02-16
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    Yep, we always have $\text{sup}(A) \in \overline{A}$, for any subset of an ordered space, as long as $\text{sup}(A)$ exists- this is a good exercise for you to try. And I don't understand your last sentence, since $C$ being closed is the same as the complement of $C$ being open. – 2017-02-17