# Topology on the set of ordinal numbers

This is a problem I encountered while reading Topology : An Outline for a First Course by Lewis E. Ward. Suppose $\Omega$ denotes the smallest ordinal number with uncountably many predecessors. Let $\mathcal{O}(\Omega)$ denote the space of ordinal numbers less than or equal to $\Omega$ endowed with the order topology.

Now, it is easy to see that $\mathcal{O}(\Omega)$ and $\mathcal{O}(\Omega) \setminus \{\Omega\}$ are normal spaces. The book says that $(\mathcal{O}(\Omega) \setminus \{\Omega\}) \times \mathcal{O}(\Omega)$ is not a normal space by giving the example of the 2 closed subsets:

1) $A:=(\mathcal{O}(\Omega)\setminus \{\Omega\}) \times \{\Omega\}$

2) $B:=\{(x,x) : x \in \mathcal{O}(\Omega) \setminus \{\Omega\} \}$

So, I need to prove that we can not find disjoint open sets $U$ and $V$ containing $A$ and $B$ respectively. I have 2 issues now:

1) I have not been able to prove the above. Can anyone tell me how to prove it?

2) While trying to prove the above, I have managed to 'disprove' the claim. That is, I think I have found sets $U$ and $V$ which satisfy the required property. So, could you tell me where I am going wrong? The construction of $U$ and $V$ is as follows:

a) Construction of $U$: For any $x \in \mathcal{O}(\Omega) \setminus \{\Omega\}$, by the well ordering principle, there exists a positive integer $n_x$, such that $\omega^{n_x-1} < x \leq \omega^{n_x}$. Then, let $$U=\cup ([x,\omega^{n_x}] \times (\omega^{n_x},\Omega])$$ where the union is over all the elements of $\mathcal{O}(\Omega) \setminus \{\Omega\}$.

b) Construction of $V$: For any $x \in \mathcal{O}(\Omega) \setminus \{\Omega\}$, by the well ordering principle, there exists a positive integer $n_x$, such that $\omega^{n_x-1} < x \leq \omega^{n_x}$. Then, let $$V=\cup ([x,\omega^{n_x}] \times [x,\omega^{n_x}])$$ where the union is over all the elements of $\mathcal{O}(\Omega) \setminus \{\Omega\}$.

• Do you know Fodor's lemma (the Pressing Down lemma)? Mar 20, 2016 at 21:10
• @AsafKaragila No Mar 20, 2016 at 21:29
• In your definition of $U$, why is $U$ open? If $x$ is a limit ordinal, I don't think $[x,\omega_x]$ is open, so there is no reason that $[x,\omega_x]\times (\omega_x, \Omega])$ is open. Am I missing something? (Same issue with $V$). Mar 21, 2016 at 0:46
• @JasonDeVito If $x$ is a limit ordinal, $\omega_x=x$. We'll have some $y<x$ such that $y$ is not a limit ordinal, and the union consists of $[y,x] \times (x,\Omega]$ and this going to be open. So, the entire union will turn out to be open. Mar 21, 2016 at 6:14
• @MathManiac: I agree there is a $y < x$ with $y$ not a limit ordinal, but I don't think there is any reason that there should be a $y<x$ with $\omega_y =x$. For example, suppose $x = \omega_\omega$. Then any $y < x$ is smaller that $\omega_{n}$ for some natural number $n$ (which can depend on $y$). Then $\omega_y \leq \omega_n < x$. Am I still missing something obvious? Mar 21, 2016 at 14:16

HINT: The first step is to prove a weak version of the pressing-down lemma.

For brevity let me write $X$ instead of $\mathcal{O}(\Omega)$. Suppose that $\varphi:X\to X$ is such that $\varphi(x)<x$ for each $x\in X$. For each $x\in X$ let $R(x)=\{y\in X:\varphi(y)\le x\}$; I claim that there is a $z\in X$ such that $R(z)$ is uncountable.

If not, for each $x\in X$ there is a $\psi(x)\in X$ such that $\psi(x)>x$, and $y<\psi(x)$ for all $y\in R(x)$. In other words, if $\varphi(y)\le x$, then $y<\psi(x)$. Let $x_0\in X$ be arbitrary. Given $x_n\in X$ for some $n\in\omega$, let $x_{n+1}=\psi(x_n)$. The sequence $\langle x_n:n\in\omega\rangle$ is an increasing sequence in $X$, so it converges to some $x\in X$. For each $n\in\omega$ we have $\psi(x_n)=x_{n+1}<x$, so $x\notin R(x_n)$, and therefore $\varphi(x)\ge x_n$. But then $\varphi(x)\ge\sup_nx_n=x>\varphi(x)$, which is absurd. This proves the claim.

Now use this and the definition of the product topology to show that if $U$ is an open nbhd of your set $A$, then there is a $z\in X$ such that

$$[z,\Omega)\times[z,\Omega)\subseteq U\;,$$

and observe that if $V$ is any open nbhd of $B$,

$$V\cap\big([z,\Omega)\times[z,\Omega)\big)\ne\varnothing\;.$$

• Seems good except a minor point. You have claimed that the sequence $x_n$ converges to some $x$. I haven't read much about ordinal numbers, so can you please tell me about a source where I could read up on why this is true? Mar 22, 2016 at 7:33
• @MathManiac: $\{x_n:n\in\omega\}$ is countable, so it has a supremum $x$. If $y<x$, then $y$ is not an upper bound for the sequence, so there is an $m\in\omega$ such that $y<x_m$. But the sequence is increasing, so $x_n\in(y,x]$ for each $n\ge m$, and it follows that the sequence converges to $x$. Mar 22, 2016 at 10:53
• And why does a countable set necessarily have a supremum? Is it obvious? Mar 22, 2016 at 14:14
• @MathManiac: Fairly obvious, yes, or at least fairly easy. Let $A\subseteq X$ be countable, and for each $x\in X$ let $P(x)=\{y\in X:y\le x\}$. Then $\bigcup_{a\in A}P(a)$ is a union of countably many countable sets, so it’s countable, and $X\setminus\bigcup_{a\in A}P(a)\ne\varnothing$. Clearly any element of $X\setminus\bigcup_{a\in A}P(a)\ne\varnothing$ is an upper bound for $A$, so $A$ has an upper bound, and the well-ordering property ensures that it has a least upper bound. Mar 22, 2016 at 14:19
• Ah, that is pretty obvious, I have used that kind of an argument before. Thanks! Mar 22, 2016 at 15:15