Checking separation axioms against topological sub-basis

Question

This is in sorts a follow up to the question Checking separation axioms against topological basis, in which I verified that checking certain separation axioms can be done only with Basis elements. My natural follow-up question is whether we can check these properties against a sub-basis instead, and I thought that it deserves a sperate thread. I think this is true but would appreciate verification for example for the following argument:

Let $(X,\tau)$ be a topological basis and let $\mathcal{S}$ be a sub-basis for $\tau$. Then $(X,\tau)$ is Hausdorff, or $T_2$, if and only if for every distinct $x,y\in X$ there exist $S_1,S_2\in \mathcal{S}$ such that $x\in S_1$, $y\in S_2$ and $S_1\cap S_2=\emptyset$.

Since $\mathcal{S}\subset \tau$, one implication is trivial. For the other direction, denote $\mathcal{B}$ the basis generated by $\mathcal{S}$. By the other question, we know that there exist $B_1,B_2\in \mathcal{B}$ such that

$$ x\in B_1, \quad y\in B_2 \quad \text{and} \quad B_1\cap B_2=\emptyset. $$

We know that the elements in $\mathcal{B}$ are finite intersections of elements in $\mathcal{S}$, and therefore

$$ B_1= \cap_{i=1}^{r_1} S_i^1 \quad \text{and} \quad B_1= \cap_{j=1}^{r_2} S_j^2. $$

Hence $x\in S_i^1$ for all $i\in [r_1]$, $y\in S_j^2$ for all $j\in [r_2]$, there exists $i_0\in [r_1]$ such that $y\notin S_{i_0}^1$ and there exists $j_0\in [r_2]$ such that $x\notin S_{j_0}^2$. So if we choose $S_1:=S_{i_0}^1$ and $S_2:=S_{j_0}^2$, we get our needed separating sets.

Does this argument seem right? It seems to work similarly well for $T_1$ and $T_0$ because we only need to concern ourselves with one open neighbourhood.

Answer 1

You have just found subbasic sets $S_1$ and $S_2$ such that $x ∈ S_1 ∌ y$ and $y ∈ S_2 ∌ x$, but the sets may not be disjoint. In fact, consider a finite discrete space – the co-singleton sets form a subbasis, but every two members intersect (if the space has at least $3$ points).

Answer 2

It will work for $T_1$ and $T_0$ but not for $T_2$ as shown by @user87690. If we have some open $O$ containing $x$ but not $y$, we can always choose it to be basic and hence of the form $\bigcap_{i=1}^n S_i$ for subbasic $S_i$; one of them does not contain $y$ and all of them contain $x$, so there also is some subbasic element to do it. This works both for $T_1$ and $T_0$.

The only useful things checkable by subbase are continuity (as inverse images preserve intersections and unions) and compactness (Alexander's subbase lemma). Convergence also won't work, e.g. But the compactness is very useful.