Separation axioms on a finite space imply discrete topology?

Question

I saw somewhere on this site, a claim that the only topology on a mertizable finite space is the discrete one. I think this stems even more generally from such a space being Hausdorff. The strongest version of which I am sure is that when the space is finite and $T_1$ then the topology has to be discrete, but I was struggling with deciding whether this is also true when the space is finite and $T_0$. I have the following counter-example:

Is the example $X=\{1,2 \}$ and $T=\Big\{ \emptyset,\{1\}, X \Big\}$ valid or am I missing something? Can this 'proposition' still be generalized using some seperation axiom?

Answer 1

You rediscovered a famous finite space, Sierpiński space, usually $X=\{0,1\}$ with topology $\{\emptyset,\{0\}, X\}$, which is the standard example of a $T_0$ but not $T_1$ space. All $T_0$ spaces are homeomorphic to subspaces of products of Sierpiński space. It’s also both an example of an included point topology (a subset is open iff it’s empty or contains $0$) and an excluded point topology (a set is open iff it’s the whole space or does not contain $1$) both of which give more general examples of such spaces.

I’m not aware of any currently studied intermediate separation axioms that are stronger than $T_0$ but weaker than $T_1$ that for finite spaces would reduce to discreteness. IIRC there is a class of finite $T_{1\over 2}$ spaces that are not discrete. The Sierpiński one being a case. (thanks to the commenter below).

Answer 2

There are some weird separation axioms weaker than $T_1$, but none of the ones I know imply discreteness on finite spaces (an exercise) [added: but see the update]:

• As Henno Brandsma mentioned, there are $T_{\frac{1}{2}}$ spaces defined by the condition that every singleton is closed or open.
• There is a notion of $_αT_1$: the $α$-topology is $T_1$ (the $α$-topology is the refinement of the original topology where you make all nowhere dense sets closed). This is equivalent to the condition that every singleton in the original topology is closed or nowhere dense.
• There is a notion of $T_{\frac{3}{4}}$, which is equivalent to the condition that every singleton is closed or regular open.

Update: there is a point-free variant of $T_1$ called subfit: a topological space $X$ is subfit if for every open sets $U ⊈ V$ there is an open set $W$ such that $U ∪ W = X ≠ U ∪ V$. Equivalently, for every point and its neighborhood $x ∈ U$ there is $y ∈ \overline{\{x\}}$ such that $\overline{\{y\}} ⊆ U$. On its own, it does not imply $T_0$, but when combined with $T_0$, it is a property we are looking for.

But first, some more context. $X$ is symmetic if for every $x ∈ U ∌ y$ ($U$ open) there is open set $V$ such that $x ∉ V ∋ y$. So topological space is $T_1$ if and only if it is $T_0$ and symmetric. Every symmetric space is subfit (it is enough to put $y = x$ in the definition of subfit). In fact, symmetric is the same as hereditarily subfit.

1. Every $T_1$ space is $T_0$ subfit.
2. Every finite $T_0$ subfit space is $T_1$ and so discrete (in fact, every finitely generated subfit space is symmetric).
3. There is a $T_0$ subfit space that is not $T_1$: take infinite set $X$ and fix a point $x ∈ X$. A set $U$ is open if and only if $x ∈ U$ and $X \setminus U$ is finite (or $U = ∅$). In other words, all points but $x$ are closed; $x$ is dense, and its neighborhood filter consists of cofinite sets.