T0 Kolmogorov Topological Spaces

We will now begin to look at a bunch of special types of topological spaces. The first of these types of topological spaces we define are called T₀ spaces, or Kolmogorov 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. $\blacksquare$