Deducing two points are topologically indistinguishable by sub-base

Question

I have the following argument which I would like to verify:

Let $X$ be a topological space with a sub-base $\{ S_\alpha \}_\alpha$ such that $\cup S_\alpha\neq X$. Then every $x,y\in X\setminus \cup S_\alpha$ are topologically indistinguishable.

And when I say topologically indistinguishable, I mean that for any open set $U$, $x,y\in U$ or $x,y\notin U$. My argument goes as follows:

Conisder the collection $\mathcal{S}:=\{ S_\alpha \}_\alpha \cup\{ X\}$. Then it generates a basis $\mathcal{B}$ whose elements are finite intersections of sets in $\mathcal{S}$. For any $x,y\in X\setminus \cup S_\alpha$ be distinct points, the only open set which contains $x$ or $y$ has to be $X$. Let $B\in \mathcal{B}$, i.e.:

$$ B=\cap_{i=1}^n S_{\alpha_i} $$

If $x\in B$, then $x\in S_{\alpha_i}$ for all $i\in [n]$, and hence $S_{\alpha_i}=X$ for all $i\in [n]$, and $B=X$. Since $\mathcal{B}$ is a basis, any open set containing $x$ must be the whole space $X$. And similarly for $y$.

This would imply that if a topological space has a sub-base which is not a cover, then it can not be a $T_0$ space. Is this reasoning sound?

Answer 1

A slightly more direct argument, avoiding your enlarged $\mathcal{S}$: So for me $\mathcal{S}$ is the given subbase for the topology on $X$, and now suppose $x \in X\setminus \bigcup \mathcal{S}$ and suppose $B$ is a basic open set that contains $x$. By definition the subbase consists of the set of all finite intersections from $\mathcal{S}$ and so $B=\bigcap \mathcal{S}'$ where $\mathcal{S}' \subseteq \mathcal{S}$ is finite. If $\mathcal{S}'$ were non-empty, say we have some $S \in \mathcal{S}'$, then $x \in S$, contradicting its choice outside the union. So $\mathcal{S'}=\emptyset$ and so $B= \bigcap \mathcal{S'}=X$. So $X$ is the only open neighbourhood of $x$.

So if the union misses more than one point, the resulting space is not $T_0$. Otherwise it will depend on the other subbase elements. E.g. Sierpiński space is $T_0$ and has subbase $\{0\}$ on the set $\{0,1\}$. The empty subbase always yields the indiscrete topology $\{\emptyset,X\}$.