T0 Kolmogorov Topological Spaces
 Definition: A topological space $X$ is said to be a T0 Space or a Kolmogorov Space if for every pair of distinct points $x, y \in X$, $x \neq y$ there exists open neighbourhoods $U$ of $x$ and $V$ of $y$ such that either $x \not \in V$ or $y \not \in U$. T0 spaces are the weakest form of separation. Many of the topological spaces that we have already looked at are in fact T0 spaces. For an example of a topological space that is not a T0 space, let $X = \{ a, b, c, d \}$ and give $X$ the indiscrete topology, i.e., $\tau = \{ \emptyset, X \}$. Then $(X, \tau)$ is not a T0 space because if we take any two distinct points $x, y \in X$, then the only open neighbourhood of $x$ is $X$, and the only open neighbourhood of $y$ is $X$ and hence $X$ cannot be a T0 space.
 Proposition: A topological space $X$ is a T0 space if and only if for every pair of distinct points $x, y \in X$, x \neq y $]], there exists an open set$U$that contains exactly one of$x$or$y$. • Proof:$\Rightarrow$Suppose that$X$is a T0 space and let$x, y \in X$,$x \neq y$. Then there exists open neighbourhoods$U$of$x$and$V$of$y$such that$x \in U$,$y \in V$, and either$x \not \in V$or$y \not \in U$. • If$x \not \in V$then$V$is an open set that contains only$y$. If$y \not \in U$, then$U$is an open set that contains only$x$. •$\Leftarrow$Suppose that for every pair of distinct points$x, y \in X$,$x \neq y$there exists an open set$U$that contains exactly one of$x$or$y$. Suppose that$U$contains only$x$. Then there must exist an open set$V$of$y$that contains only$y$(otherwise every open set containing$y$contains$x$- a contradiction). So$X$is a T0 space.$\blacksquare\$