Closed disjoint sets in $\omega_1$ implies one of them is countable

Question

Let $A, B \subseteq \omega_1$ be disjoint, closed sets. Show that one of A or B must be countable.

Since $\omega_1$ is uncountable, at least one of A or B must be countable, I want to show that exactly one of A or B must be countable.

I think my answer is too simple, so there must be a flaw somewhere. Let B be the set that contains $min(\omega_1)$ take min(A), this must exist since A is a subset of a well-order hence, is a well order itself, since we took min(A) = a, pred(a) = B which implies B should be countable. where pred(a) = $\{x \in \omega_1 : x \leq a \}$

Edit: I realize that pred(A) $\subset$ B, so it doesn't really show that one of the sets is countable. So second attempt, if they are disjoint one of them must contain $min(\omega_1)$ WLOG let B contain the minimum. Lets suppose both A and B are uncountable, B will contain a tail of the pred(a) for $\forall a \in A$. I'm not sure how to continue from here, both of them are closed so contain their limit point, I want to show that if they are both uncountable they will have sequences that converge to the same point which will give a contradiction.. Let B contain pred(a) for a = min(A). So the entire pred(a) is contained in B, pred(a) is countable, we can construct a sequence in B s.t for any elemnts $a_n$ of pred(a) there exists $b_n$ with the property $a_n \leq b_n \leq a_{n+1}$ for all n (we can do this because B is uncountable), then both sequences converge to min(a) which is a contradiction. So B must be countable.

Answer 1

First, $\min\omega_1=0$, and there is no reason to suppose that it belongs to either of the sets $A$ and $B$. Secondly, it’s not true that exactly one of $A$ and $B$ is countable: they could both be finite, one could be countably infinite and the other finite, or both could be countably infinite. (For example, $A$ could be $\{a\}$ for some $a\in\omega_1$, and $B$ could be $\{b\}$ for any $b\in\omega_1\setminus A$.) All that you can show is that if $A$ and $B$ are disjoint, closed subsets of $\omega_1$, then at least one of them is countable.

Suppose that $A$ and $B$ are uncountable closed subsets of $\omega_1$; we’ll show that they are not disjoint. Let $a_0\in A$ be arbitrary, and let $B_0=\{b\in B:a_0<b\}$.

• Show that $B_0$ is uncountable and hence non-empty.

Let $b_0=\min B_0$, and let $A_0=\{a\in A:b_0<a\}$.

• Show that $A_0$ is uncountable and hence non-empty.

Let $a_1=\min A_0$.

In general, suppose that $n\in\Bbb N$, and we’ve chosen $a_k$ for $k\le n$ and $b_k$ for $k<n$. Let $$B_n=\{b\in B:a_n<b\}\;,$$ show that $B_n\ne\varnothing$, let $b_n=\min B_n$, let $$A_{n+1}=\{a\in A:b_n<a\}\,$$ show that $A_{n+1}\ne\varnothing$, and let $a_{n+1}=\min A_{n+1}$. Now we have $a_k$ for $k\le n+1$ and $b_k$ for $b<n$, so the recursive construction can be continued for all $n\in\Bbb N$.

At this point you have two sequences, $\langle a_n:n\in\Bbb N\rangle$ and $\langle b_n:n\in\Bbb N\rangle$, the first in $A$ and the second in $B$, and they’re interlaced:

$$a_0<b_0<a_1<b_1<a_2<b_2<\ldots a_n<b_n<a_{n+1}<\ldots$$

• Show that $a=\sup_na_n$ exists. Use the fact that $A$ is closed to show that $a\in A$.
• Similarly, show that $b=\sup_nb_n$ exists and that $b\in B$.
• Show that $a=b$, and conclude that $A\cap B\ne\varnothing$.