Prove: If $A$ and $B$ are closed subsets of $[0,\Omega]$ then at least $A$ or $B$ is bounded

As usual, I am self studying topology and my knowledge of ordinals is meagre. Have done some research on it.

Theorem 5.1 Any countable subset of $[0,\Omega)$ is bounded above.

(This exercise requires a knowledge of ordinals) From problems:

1)A countable space need not be compact

2)countable compact space is pseudocompact

it is immediate that the the space of countable ordinals is pseudocompact. In fact, it actually more than just pseudocompact. In fact, it is actually more than just pseudocompact because a continuous real valued function on $[0,\Omega)$ is ultimately constant ,so is more than just bounded. (A function $f$ on $[0, \Omega)$ if there is an ordinal number $a\in [0,\Omega)$ such that $f(x)=f(y)$ $\forall x,y\in(a, \Omega)$.) Showing that every continuous real valued function on [0, $\Omega$) is ultimately continuous is not easy. Use the following out which is adapted from chapter 5 and 6 of Gillman and Jerison[6]

Note: I don’t have this book.

(a) Claim If A and B are closed sets of $[0,\Omega)$, then atleast A or B is bounded.

Attempted proof

Let A and B be closed subsets of $[0,\Omega)$ Then there is an ordinal number $a<\Omega$ such that $A \in [0,a]$ or $B\in [0,a]$. Since $A$ and $B$ are infinite, it contains a cluster point. So A or B is countably compact So by Theorem 5.1 is bounded.

I think my first effort sucks. On second thought, I thought of following something like this argument closed intervals of $\omega_1$ are compact to do it.

Any help to solve this theorem would be appreciated.

Solution 1:

If both $A$ and $B$ are (closed and) unbounded pick $a_0 \in A$ and $b_0 \in B$ with $a_0 < b_0$, and next $a_1 \in A$ with $a_1 > b_0$ and so on till we have interspersed sequences $a_0 < a_1 < a_2 < \ldots$ in $A$ and $b_0 < b_1 < b_2 < \ldots$ in $B$ with $a_n < b_n$ for all $n$. Then define $\alpha=\sup_n a_n \in [0,\Omega)$ and also $\beta=\sup_n b_n \in [0,\Omega)$ and by closedness of $A$ and $B$ (and the intersepresedness) we have $\alpha=\beta\in A \cap B$ contradicting their disjointness.

We only need that a sequence of countable ordinals has a sup that is still a countable ordinal.

It's easy that $X=[0,\Omega)$ is countably compact, just use sup argument for a countable subset of $X$ and it's clear that $X$ is not compact, as all initial segements for a cover without a finite (or even countable) subcover.

Pseudocompactness follows from countable compactness and needs no separate argument.