The particular-point topology is $T_0$, it is not $T_1$ and not regular once the set has at least two points, and it is not normal once the set has at least three

Let $X$ be a set, fix $p \in X$, and give $X$ the particular-point topology $\mathcal{T}_p = \{\varnothing\} \cup \{\, U \subseteq X : p \in U \,\}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), whose closed sets are $X$ together with the subsets of $X$ that do not contain $p$. Then:

1. $(X, \mathcal{T}_p)$ is $T_0$ ($T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces), for every $X$ and every $p$.
2. If $X$ has at least two points then $(X,\mathcal{T}_p)$ is not $T_1$, and hence not Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).
3. If $X$ has at least two points then $(X,\mathcal{T}_p)$ is not regular (Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly).
4. If $X$ has at least three points then $(X,\mathcal{T}_p)$ is not normal (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly).

Clause 4 needs one point more than clause 3, and the extra point is not slack: on a two-point set the particular-point topology is Sierpinski space, which is normal (Sierpinski space is $T_0$ and normal but neither $T_1$ nor regular: normality without $T_1$ implies nothing). So this family separates the two failures, and it shows that "not regular" and "not normal" begin at different sizes.

• Closures here are as large as they can be. For $A \ne \varnothing$ with $p \notin A$ one has $\overline{A} = A$, since $A$ is already closed; but $\overline{\{p\}} = X$, because the only closed set containing $p$ is $X$ (A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$; the closure is the smallest closed superset and equals $A$ together with its derived set). So the particular point is dense, and a single point can be dense in a space with any number of points at all.

• Every nonempty open set contains $p$, which is the one fact behind clauses 3 and 4 alike. It makes the space as far from Hausdorff as possible while still distinguishing points: no two nonempty open sets are ever disjoint.

• Sierpinski space is the case of two points. With $X = \{p, x\}$ the open sets are $\varnothing$, $\{p\}$ and $X$, which is The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies's Sierpinski topology with the open point named $p$; so clause 4 genuinely needs its extra hypothesis.