Example of a T2 space that is not T3.

Question

I am currently solving a problem where I need to bring up a counterexample in the form of a T2 space that is not T3. The Wikipedia page of T3 brings an example that states: " There exist Hausdorff spaces that are not regular. An example is the set $\mathbb{R}$ with the topology generated by sets of the form $U\backslash C$, where $U$ is an open set in the usual sense, and $C$ is a fixed nonclosed subset of R with empty interior. "

Let $C = \left\lbrace \frac{1}{n} : n \in \mathbb{N}\right\rbrace$, what would be a closed set $K$ and a point $p$ such that there is no $U_K\supset K$ and $U_x\supset\lbrace x\rbrace$ open with $U_K\cap U_x\neq\varnothing$?

Is there something I am missing?

Thank you in advance for your answers.

Edit 1:

Reading again the example given by Wikipedia it seems like it isn't even T2! Any point in $C$ is never contained in any open set!

Edit 2:

The Wikipedia Page has now been fixed. Thank you @PatrickR.

Answer 1

Adding this for clarification.

There may be a mistake. Let's denote your space $\mathbb{R}_C$. Sets that are open in $\mathbb{R}$ (with the standard topology) should also be open in $\mathbb{R}_C$. Note that this topology is strictly finer than $\mathbb{R}$.

It is easy to check that $\mathbb{R}_C$ is Hausdorff by the following straightforward fact:

If $(X, \mathcal{T})$ is Hausdorff and $(X, \mathcal{T}')$ is finer (i.e., $\mathcal{T} \subseteq \mathcal{T}'$), then $(X, \mathcal{T}')$ is Hausdorff.

To see that $\mathbb{R}_C$ is not $T_3$ (regular (Hausdorff)), consider the obvious point $p = 0$ and the closed set $C$ (which misses $p$). We claim that there is no neighborhood of $p$ disjoint from an open set containing $C$. Consider a basis element containing $p$. It must be of the form $B'_\varepsilon = (-\varepsilon, \varepsilon) - C$, where $\varepsilon > 0$ (otherwise the basis element intersects $C$). Now, by Archimedean property, there is some $n \in \mathbb{Z}^+$ such that $n > \frac{1}{\varepsilon} \implies \frac{1}{n} < \varepsilon$. Consider a basis element about $\frac{1}{n} \in C$; it must be of the form $B_\delta = \left( \frac{1}{n} - \delta, \frac{1}{n} + \delta \right)$, where $\delta > 0$. If $\delta \le \frac{1}{n} - \frac{1}{n + 1} = \frac{1}{n (n + 1)}$, then the point $\frac{1}{n} - \frac{1}{2} \delta \in B'_\varepsilon \cap B_\delta$. Otherwise, the point $\frac{1}{2} \left( \frac{1}{n} + \frac{1}{n + 1} \right) \in B'_\varepsilon \cap B_\delta$. So, $p$ cannot be separated from $C$ by open sets.

N.B. This is actually a well-known topology, and it is called the $K$-topology on $\mathbb{R}$. (The set $C$ you defined is called $K$.)

Answer 2

For examples, look at pi-base

Filter by formula "Hausdorff + ~Regular" to find many of these examples.

Answer 3

Instead of considering the topology generated by the sets of the form $U \setminus C$, with $U$ an usual open and $C = \{\frac{1}{n} : n \in \mathbb{N}\}$, consider the topology generated by the sets of the form $U \setminus V$, with $U$ an usual open and $V \subseteq C = \{\frac{1}{n} : n \in \mathbb{N}\}$. (See the definition here: https://topology.pi-base.org/spaces/S000056).

Let $U$ be an usual open, $U$ is open in this topology too ($U = U \setminus \emptyset$). Since this topology is finer than the usual and the usual is Hausdorff, this one is Hausdorff too.

But $C$ is closed in this topology (it is clear that $\mathbb{R} \setminus C$ is open by definition of this topology), $0 \notin C$, and $0$ and $C$ can't be separated by disjoint neigborhoods.