Why is a Countable Basis Needed in This Proof?

Question

My question is regarding the Theorem in Munkres that states: Every Regular Space with a Countable Basis is Normal. Before reading the proof in Munkres, I tried to prove it myself and came up with a "proof" that is clearly wrong, but I am not exactly sure why it fails. Here is my attempt: Let $X$ be a Regular topological space with a countable basis. Consider two disjoint closed sets $C, D$. Then $\forall x \in C$, use regularity to find disjoint open neighborhoods $U_x, V$ such that $x \in U$ and $D \subset V$. Then $\bigcup U_x \forall x \in C$ is an open set containing $C$ that is disjoint from $V$.

I know this proof is incorrect because it never uses the fact that $X$ has a countable basis and would imply that all regular spaces are normal, which we know is not true. However, I am having a difficult time seeing where exactly my proof fails. My thought is that it has to do with the idea of choosing an open $ U_x \forall x \in C$. I fear this may not be possible in a case where there are uncountable many elements in $C$. However, I would think that the Axiom of Choice would allow me to make such a statement. So, where exactly is my logic failing here?

Answer 1

(To remove from unanswered)

As commented your proof attempt doesn't work because choosing $\bigcup_x U_x$ as your cover of $C$ forces you to choose $\bigcap_x V_x$ as your cover of $D$, which needn't be open. In fact, even if you replace those unions/intersections by ones of countably infinite size (which the second countable or Lindelof hypothesis allows you to do), $\bigcap_x V_x$ might not be open. We can get a sharper result by a cleverer choice of cover of $C$.

In my view the "real" reason for this is that countable cardinals have the following special (but trivial) property: (i) they can be "reached" from finite cardinals, as the union of all those lesser than themselves and (ii) any topology is closed under finite intersections.

The usual proof that Lindelof + Regular $\to$ normal (observe that second countable spaces are Lindelof) goes as follows:

Take disjoint closed (nonempty) $A$ and $B$. For $a\in A$ there are $U_a$ and $V_a$ disjoint open neighbourhoods of $a$ and $B$ respectively. Since $A$ inherits the Lindelof property (it is closed) there is a countable $F\subseteq A$ with $A\subseteq\bigcup_{a\in F}U_a$. We take a surjective function $\phi:\omega\to F$ which is either a bijection or is an injection up to some point $n<\omega$ beyond which $\phi$ is constant. If $U'_k:=\bigcap_{j\le k}U_{\phi(k)}\setminus\overline{V_{\phi(j)}}$, $V'_k:=\bigcap_{j\le k}V_{\phi(k)}\setminus\overline{U_{\phi(j)}}$ for all (ordinals) $k<\omega$ then these are all open because the intersection of $n$-many (finite) open sets is open and $U':=\bigcup_{k<\omega}U'_k,\,V':=\bigcup_{k<\omega}V'_k$ are disjoint open sets.

Moreover, $U$ covers $A$ and $V$ covers $B$ by an easy check. Thus the space is normal.

We essentially approximate $\bigcup_{a\in F}U_a$ but remove the bits that stray too close to $B$. The only point at which countability of $F$ is used is when we use the fact that for any ordinal (equivalently, any cardinal) $n<\omega$ the intersection of $n$-many opens is always open - because $n$ is finite. The main reason I wanted to write this answer is because your question made me realise the following point: the theorem and its proof carries over verbatim to the more general situation:

If a space is regular and has a basis of cardinality $\kappa$ and has the property that if $\lambda<\kappa$ is any smaller cardinal then the intersection of $\lambda$-many open sets is open, then the space is normal.

Or:

If a space is regular and is $\kappa$-Lindelof for some cardinal $\kappa$ and for any smaller cardinal $\lambda<\kappa$ the intersection of $\lambda$-many open sets is open, then the space is normal.

This makes me feel that, in this context, the only special thing about countably infinite bases is that a topology is by definition closed under intersections of size strictly smaller than "countably infinite" (i.e. finite). If you have a space satisfying a stronger closure-under-intersection property, the same proof idea will work. These stronger properties are met by some spaces, so this isn't a completely pointless observation; for example, giving $\Bbb R$ its cocountable topology, the intersection of $\omega$-many open sets is always open. Therefore the above theorem would say, if $\Bbb R$ has (in its cocountable topology) a basis of size $\omega_1=\aleph_1$ and if $\Bbb R$ is regular then it will be normal. In this particular instance it doesn't work because the space fails to be regular (and my set theory isn't strong enough to say at a glance whether or not it does have a basis of size $\aleph_1$).