T2 Hausdorff Topological Spaces
Recall from the T0 Kolmogorov Topological Spaces page that a topological space $X$ is said to be a T_{0} space or a Kolmogolov space if for every pair of distinct points $x, y \in X$ there exists open neighbourhoods $U$ of $x$ and $V$ of $y$ such that either $x \not \in V$ or $y \not \in U$.
Also recall from the T1 Fréchet Topological Spaces page that a topological space $X$ is said to be a T_{1} space or a Fréchet space if for every pair of distinct points $x, y \in X$ there exists open neighbourhoods $U$ of $x$ and $V$ of $y$ such that $x \not \in V$ and $y \not \in U$.
We will now revisit a concept that we have already looked at before: Hausdorff spaces.
Definition: A topological space $X$ is said to be a T_{2} Space or a Hausdorff Space if for every pair of distinct points $x, y \in X$ there exists open neighbourhoods $U$ of $x$ and $V$ of $y$ such that $U \cap V = \emptyset$. |
We have already seen many examples of T_{2} spaces. Of course, there exists some spaces that are T_{1} and not T_{2} though. For example, consider any infinite set $X$ with the cofinite topology:
(1)Let $x, y \in X$ with $x \neq y$. Consider the sets $U = X \setminus \{ y \}$ and $V = Y \setminus \{ x \}$. Then $U$ and $V$ are open neighbourhoods of $x$ and $y$ respectively and notice that $x \not \in V$ and $y \not \in U$. So $X$ is a T_{1} space.
However, we claim that $X$ is not a T_{2} space. Let $x, y \in X$ with $x \neq y$. Then any open neighbourhood $U$ of $x$ is of the form:
(2)Where $x \neq x_i$ for all $i \in \{1, 2, ..., n \}$. Similarly, any open neighbourhood $V$ of $y$ is of the form:
(3)Where $y \neq y_j$ for all $j \in \{1, 2, ..., m \}$. Note that we always have that $U \cap V \neq \emptyset$ though! So $X$ cannot be a T_{2} space.
So we have established that there exists T_{1} spaces that are not T_{2}. As expected, every T_{2} space is a T_{1} space though.
Theorem 1: Let $X$ be a topological space. If $X$ is a T_{2} space then $X$ is a T_{1} space. |
- Proof: Let $X$ be a T_{2} space and let $x, y \in X$ where $x \neq y$. Since $X$ is a T_{2} space there exists open neighbourhoods $U$ of $x$ and $V$ of $y$ such that $U \cap V = \emptyset$. Then these such neighbourhoods satisfy "$x \not \in V$ and $y \not \in U$.
- So $X$ is a T_{1} space. $\blacksquare$