Tychonov vs Hausdorf separation property

Question

I look for an example of a topology which is Tychonov (T1) but not Hausdorf (T2). In this case the neighborhoods for any two points x and y must intersect at, say, point z. If we consider three-points space with topology T=$\{\emptyset,\{x,z\}.\{y,z\},\{x,y,z\}\}$, then points x and z are not separated in T1 sense. How to modify this example to have T1 without T2? Thank you.

Answer 1

A Tychonov space is not the same as a $T_1$ space in normal topological usage. I think Fréchet was the first two assume closedness of finite sets (which is what $T_1$ means) as an axioms for topologies.

A finite $T_1$ space has all its subsets closed, so all its subsets open,and so is discrete and trivially Hausdorff.

Your example is not a topology, as $\{x,z\} \cap \{y,z\} = \{z\}$ is not in the topology.

A satndard example for your question is to take any inifnite set, say $X =\mathbb{N}$ and declare as closed sets all finite ones (including the empty set) and $X$ itself. The open sets are just the complements. This is $T_1$ by construction and one can show that any two non-empty open sets have non-empty intersection, so $X$ is not Hausdorff in this topology.