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 $\mathcal{O}(\Omega) \setminus \{\Omega\}$.

  • $\begingroup$ Do you know Fodor's lemma (the Pressing Down lemma)? $\endgroup$
    – Asaf Karagila
    Mar 20, 2016 at 21:10
  • $\begingroup$ @AsafKaragila No $\endgroup$
    – MathManiac
    Mar 20, 2016 at 21:29
  • $\begingroup$ 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 $V$). $\endgroup$ Mar 21, 2016 at 0:46
  • $\begingroup$ @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. $\endgroup$
    – MathManiac
    Mar 21, 2016 at 6:14
  • 1
    $\begingroup$ @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? $\endgroup$ Mar 21, 2016 at 14:16

1 Answer 1


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 $B$,


  • $\begingroup$ 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? $\endgroup$
    – MathManiac
    Mar 22, 2016 at 7:33
  • $\begingroup$ @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 $x$. $\endgroup$ Mar 22, 2016 at 10:53
  • $\begingroup$ And why does a countable set necessarily have a supremum? Is it obvious? $\endgroup$
    – MathManiac
    Mar 22, 2016 at 14:14
  • $\begingroup$ @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. $\endgroup$ Mar 22, 2016 at 14:19
  • $\begingroup$ Ah, that is pretty obvious, I have used that kind of an argument before. Thanks! $\endgroup$
    – MathManiac
    Mar 22, 2016 at 15:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.