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 $\mathcal{O}(\Omega) \setminus \{\Omega\}$.
 A: 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 $B$,
$$V\cap\big([z,\Omega)\times[z,\Omega)\big)\ne\varnothing\;.$$
