A T2 Space That Is NOT a T3 Space

A T2 Space That Is NOT a T3 Space

Recall from the T3 (Regular Hausdorff) Topological Spaces page that a topological space $X$ is said to be regular if for all $x \in X$ and all closed sets $F$ of $X$ not containing $x$ there exists open sets $U$ and $V$ that separate $\{ x \}$ and $F$.

We then said that a topological space $X$ is a T3 space if it is both a regular space and a T1 space.

We saw that every T3 space is a T2 space. We will now see that there exists T2 spaces that are not T3 spaces.

Consider the following half-plane:

\begin{align} \quad X = \{ (x, y) \in \mathbb{R}^2 : y \geq 0 \} \end{align}

Define the topology on $X$ whose basis elements are open disks contained in $X$:


As well as sets which we denote by:

\begin{align} \quad U_{a, r} = B((a, 0), r) \cap X \setminus \{ (x, 0) \in \mathbb{R}^2 : x \in (a - r, a + r) \setminus \{ a \} \} \end{align}

Then $X$ with this topology is clearly T2. Any two points not on the $x$-axis can be separated by open balls. A point not on the $x$-axis can be separated from a point on the $x$-axis by taking an open ball and open half-disk with a sufficiently small radius. The same goes for two points on the $x$-axis.

However, we claim that $X$ is not a T3 space. Consider the following set:

\begin{align} \quad F = \{ (x, 0) \in \mathbb{R}^2 : -1 < x < 1 \} \end{align}

Take the point $\mathbf{x} = (1, 0)$. Then any open set containing $\mathbf{x}$ will contain a set of the form $U_{1, r}$ where $r > 0$. But since $r \neq 0$, $U_{1, r}$ will intersect any open set containing $F$ and hence $X$ is not a T3 space.

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