Let \(X\) be a topological space. Two subsets \(S, T\subseteq X\) are separated by neighborhoods if there exist disjoint open sets \(U\supseteq S\) and \(V\supseteq T\).
\(X\) is \(T_0\) if for all \(x_1\neq x_2\) in \(X\) some open set contains exactly one of \(x_1,x_2\).
\(X\) is \(T_1\) if for all \(x_1\neq x_2\) in \(X\) there is an open set containing \(x_1\) but not \(x_2\), and an open set containing \(x_2\) but not \(x_1\).
\(X\) is \(T_2\) if any two distinct points are separated by neighborhoods; this is the Hausdorff condition.
\(X\) is \(T_{2.5}\) if any two distinct points \(x_1,x_2\) have open neighborhoods \(U_1\ni x_1\) and \(U_2\ni x_2\) with \(\overline{U_1}\cap\overline{U_2}=\emptyset\).
\(X\) is \(T_3\) if it is \(T_0\) and regular: for every \(x\in X\) and every closed \(F\subseteq X\) with \(x\notin F\), the sets \(\ts{x}\) and \(F\) are separated by neighborhoods.
\(X\) is \(T_{3.5}\) if it is \(T_0\) and completely regular: for every \(x\in X\) and every closed \(F\subseteq X\) with \(x\notin F\) there is a continuous \(f\colon X\to[0,1]\) with \(f(x)=0\) and \(f(F)\subseteq\ts{1}\).
\(X\) is \(T_4\) if it is \(T_1\) and normal: any two disjoint closed subsets of \(X\) are separated by neighborhoods.
A space is \(T_1\) if and only if every one-point subset is closed.
A space that is not \(T_0\): let \(\mathcal L^2(\RR)\) be the set of measurable \(f\colon\RR\to\CC\) with \(\int_\RR\abs{f}^2<\infty\), topologized by the pseudometric \(d(f,g)=\qty{\int_\RR\abs{f-g}^2}^{1/2}\). If \(f\neq g\) agree almost everywhere, then \(d(f,g)=0\), so every open set containing one of them contains the other.
A space that is \(T_0\) but not \(T_1\): for a commutative ring \(R\) with a prime ideal that is not maximal, \(\Spec R\) with the Zariski topology. The closure of the point \(\mathfrak p\) is \(V(\mathfrak p)=\ts{\mathfrak q\in\Spec R \st \mathfrak q\supseteq\mathfrak p}\), so a prime that is not maximal is not a closed point; distinct primes have distinct closures, which gives \(T_0\).