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 T_{0} 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 T_{1}, or Fréchet spaces.
Definition: A topological space $X$ is said to be a T_{1} 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 T_{1}. 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 T_{0} and not T_{1}. A famous (and thankfully simple) example is the Sierpiński space defined by the two point set $X = \{ 0, 1 \}$ and the topology:
(1)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 T_{0} 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 T_{1}
While not every T_{0} space is T_{1}, the converse is true, i.e., every T_{1} space is a T_{0} space.
Theorem 1: Let $X$ be a topological space. If $X$ is a T_{1} space then $X$ is a T_{0} space. |
- Proof: Suppose that $X$ is a T_{1} space and let $x, y \in X$ with $x \neq y$. Then since $X$ is a T_{1} 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 T_{0} space since the condition "either $x \not \in V$ or $y \not \in U$" is satisfied.
- Therefore, $X$ is a T_{0} space. $\blacksquare$
The next result gives us a characterization for determining whether a space is T_{1} or not.
Proposition 1: Let $X$ be a topological space. Then $X$ is T_{1} if and only if every singleton set in $X$ is closed. |
- Proof: $\Rightarrow$ Suppose that $X$ is a T_{1} 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 T_{1} 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 T_{1} space. $\blacksquare$