T1 Fréchet Topological Spaces

# T1 Fréchet Topological Spaces

Recall from the T0 Kolmogorov Topological Spaces page that we said that a topological space $X$ is said to be a T0 space or a Kolmogorov space if for all pairs 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 $y \not \in U$.

We will now look at a stronger type of separated space known as T1, or Fréchet spaces.

 Definition: A topological space $X$ is said to be a T1 Space or Fréchet 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 $x \not \in V$ and $y \not \in U$.

For example, if $X = \{ a, b, c, d \}$ is given the discrete topology then $X$ is T1. This is because for any pair of points $x, y \in X$, the open neighbourhoods $U = \{ x \}$ of $x$ and $V = \{ y \}$ of $y$ are such that $x \not \in V$ and $y \not \in U$.

There are some examples of topological spaces that are T0 and not T1. A famous (and thankfully simple) example is the Sierpiński space defined by the two point set $X = \{ 0, 1 \}$ and the topology:

(1)
\begin{align} \quad \tau = \{ \emptyset, \{ 0 \}, \{ 0, 1 \} \} \end{align}

Notice that the only two points in $X$ are $0$ and $y$. If we let $U = \{ 0 \}$ be an open neighbourhood of $0$ and $V = \{ 0, 1 \}$ be the open neighbourhood of $1$, then we clearly see that $X$ is a T0 space.

However, the only open neighbourhoods of $0$ are $\{ 0 \}$ and $\{ 0, 1 \}$ and the only open neighbourhood of $1$ is $\{ 0, 1 \}$. So we must set $V = \{ 0, 1 \}$. But $U \subseteq V$ for both open neighbourhoods $U$, so $0$ is contained in $V$ always, so $X$ is not T1

While not every T0 space is T1, the converse is true, i.e., every T1 space is a T0 space.

 Theorem 1: Let $X$ be a topological space. If $X$ is a T1 space then $X$ is a T0 space.
• Proof: Suppose that $X$ is a T1 space and let $x, y \in X$ with $x \neq y$. Then since $X$ is a T1 space there exists open neighbourhoods $U$ of $x$ and $V$ of $y$ such that $x \not \in V$ and $y \not \in U$. But this means that $X$ is also a T0 space since the condition "either $x \not \in V$ or $y \not \in U$" is satisfied.
• Therefore, $X$ is a T0 space. $\blacksquare$

The next result gives us a characterization for determining whether a space is T1 or not.

 Proposition 1: Let $X$ be a topological space. Then $X$ is T1 if and only if every singleton set in $X$ is closed.
• Proof: $\Rightarrow$ Suppose that $X$ is a T1 space. Let $x \in X$ and consider the singleton set $\{ x \}$. We will show that $\{ x \}^c = X \setminus \{ x \}$ is open. Let $y \in X \setminus \{ x \}$. Then $x \neq y$. Since $X$ is a T1 space there exists an open neighbourhood $V$ of $y$ such that $x \not \in V$. But then $V \subseteq \{ x \}^c = X \setminus \{ x \}$. So $y \in \mathrm{int} (\{ x \}^c) = \mathrm{int}(X \setminus \{ x \})$ which shows that $\{ x \}^c = X \setminus \{ x \}$ is open. So each singleton set $\{ x \}$ is closed.
• $\Leftarrow$ Suppose that every singleton set in $X$ is closed. Let $x, y \in X$ and suppose that $x \neq y$. Then $\{ x \}$ and $\{ y \}$ are closed. Let $U = \{ y \}^c = X \setminus \{ y \}$ and let $V = X \setminus \{ x \}$. Then $U$ and $V$ are open neighbourhoods of $x$ and $y$ respectively such that $x \not \in V$ and $y \not \in U$. So $X$ is a T1 space. $\blacksquare$