T2 Hausdorff Topological Spaces

# T2 Hausdorff Topological Spaces

Recall from the T0 Kolmogorov Topological Spaces page that a topological space $X$ is said to be a T0 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 T1 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 T2 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 T2 spaces. Of course, there exists some spaces that are T1 and not T2 though. For example, consider any infinite set $X$ with the cofinite topology:

(1)
\begin{align} \quad \tau = \{ U \subset X : U^c \: \mathrm{is \: finite \: or \: } U = \emptyset \} \end{align}

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 T1 space.

However, we claim that $X$ is not a T2 space. Let $x, y \in X$ with $x \neq y$. Then any open neighbourhood $U$ of $x$ is of the form:

(2)
\begin{align} \quad U = X \setminus \{ x_1, x_2, ..., x_n \} \end{align}

Where $x \neq x_i$ for all $i \in \{1, 2, ..., n \}$. Similarly, any open neighbourhood $V$ of $y$ is of the form:

(3)
\begin{align} \quad V = X \setminus \{ y_1, y_2, ..., y_m \} \end{align}

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 T2 space.

So we have established that there exists T1 spaces that are not T2. As expected, every T2 space is a T1 space though.

 Theorem 1: Let $X$ be a topological space. If $X$ is a T2 space then $X$ is a T1 space.
• Proof: Let $X$ be a T2 space and let $x, y \in X$ where $x \neq y$. Since $X$ is a T2 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 T1 space. $\blacksquare$