A T3 Space That Is NOT a T4 Space

Recall from the T4 (Normal Hausdorff) Topological Spaces page that a topological space $X$ is said to be normal if for all disjoint closed sets $E$ and $F$ there exists open sets $U$ and $V$ with $E \subseteq U$, $F \subseteq V$, and $U \cap V = \emptyset$.

We then said that a topological space $X$ is a T₄ space if it is both a normal space and a T₁ space.

We saw that every T₄ space is a T₃ space. We will now see that there exists T₃ spaces that are not T₄ spaces.

Recall from The Lower and Upper Limit Topologies on the Real Numbers page that the set $\mathbb{R}$ with the topology $\tau$ generated by sets of the form $[a, b)$ where $a, b \in \mathbb{R}$, $a < b$ is called the lower limit topology (or the Sorgenfrey line). We claim that $\mathbb{R}$ with the topology is T₃.

To show this, let $x \in \mathbb{R}$ and consider any closed set $F$ in $\mathbb{R}$ with the lower limit topology. Then $\mathbb{R} \setminus F$ is an open set containing $x$. So there exists a $b \in \mathbb{R}$ such that $x \in [x, b)$ where, of course, $[x, b)$ is an open set. Let $U = [x, b)$. Then $U^c = (-\infty, x) \cup [b, \infty)$ is also an open set in $\mathbb{R}$ with the lower limit topology and it contains $F$. Let $V = (-\infty, x) \cup [b, \infty)$.

Then, for every point $x \in \mathbb{R}$ and for all closed sets $F$ in $\mathbb{R}$ not containing $X$ there exists open sets $U$ and $V$ that separate $\{ x \}$ and $F$, so $\mathbb{R}$ with the lower limit topology is a T₃ space.

Now we know that the product of two T₃ spaces is T₃, so $\mathbb{R} \times \mathbb{R} = \mathbb{R}^2$ where each copy of $\mathbb{R}$ has the lower limit topology is also a T₃ space. We claim that this product is not a T₄ space though.

Consider the following set in $\mathbb{R}^2$ with this topology:

(1)

\begin{align} \quad \{ (a, -a) \in \mathbb{R}^2 : a \in \mathbb{R} \} \end{align}

This set is closed in $\mathbb{R}^2$ with respect to this topology. Moreover, if we consider this set as a subspace of $\mathbb{R}^2$ this this topology then every singleton set is open which implies that this line has the discrete topology, i.e., every subset of points on this line is both open and closed. In particular, the following sets are closed in the subspace of this line:

(2)

\begin{align} \quad E^* = \{ (a, -a) \in \mathbb{R}^2 : a \in \mathbb{Q} \} \end{align}

(3)

\begin{align} \quad F^* = \{ (a, -a) \in \mathbb{R}^2 : a \not \in \mathbb{Q} \} \end{align}

Since these two sets are closed in this subspace, there exists closed sets $E$ and $F$ in $\mathbb{R}^2$ such that $E^* = E \cap \mathbb{R}^2$ and $F^* = F \cap \mathbb{R}^2$. Clearly $E^*$ is contained in $E$ and $F^*$ is contained in $F$. Moreover, it's not hard to see that there exists no open sets $U$ and $V$ which can separate $E$ and $F$ because overlap of $U$ and $V$ will always occur.

So $\mathbb{R}^2$ is not normal with this topology and hence $\mathbb{R}^2$ is not a T₃ space that is not a T₄ space with this topology.