Show that $A$ or $B$ must be bounded in $\omega_1$ This is a question which took me longer to figure out than I would have liked. I could not find it on this website so I thought I would share my solution here.
Let $A$ and $B$ be disjoint closed subsets of $\omega_1$, the first uncountable ordinal. Show that either $A$ or $B$ must be bounded in $\omega_1$.
I have left my solution below.
 A: Assume that both $A$ and $B$ are unbounded. Because $A$ and $B$ are unbounded, we may recursively define a strictly increasing sequence such that every other element is in $A$ or $B$ respectively. Choose some element $a_0\in A$. If $i\in\mathbb{Z}^+$ is odd, let $a_i\in B$ such that $a_{i-1}<a_i$ and if $i$ is even, let $a_i\in A$ such that $a_{i-1}<a_i$. Now, let $a=\cup_{i\in\mathbb{N}}a_i=\sup\{a_i:i\in\mathbb{N}\}$. $a$ is a countable union of countable ordinals and is therefore a countable ordinal (and thus an element of $\omega_1$).
Let $(x,y)$ be an open interval in $\omega_1$ containing $a$ so that $x<a<y$. Then $x\in a$ which means there must exist $i\in\mathbb{N}$ such that $x\in a_i$ so that $x<a_i$. Clearly $a_{i+1}<y$, for if it were not, we would have $a_{i+1}<a<y\leq a_{i+1}$. Thus, $x<a_i<a_{i+1}<y$, so $a_i$ and $a_{i+1}$ are both in $(x,y)$, which means every open interval containing $a$ must contain an element of $A$ and an element of $B$.
Since $a$ is clearly not $0$, every open set containing $a$ contains an open interval and therefore intersects both $A$ and $B$. Hence, $a$ is a limit point of both $A$ and $B$. Since $A$ and $B$ are closed, this implies $a$ is an element of both $A$ and $B$ which contradicts the fact that $A$ and $B$ are disjoint. So, by way of contradiction, either $A$ or $B$ must be bounded.
