# Example of a T2 space that is not T3.

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?

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:

• There might be a typo. The topology should also include all sets that are also open in the usual topology on $\mathbb{R}$. Commented Aug 2 at 19:29
• As discussed, the topology is not only not $T_2$, it is not even $T_0$, because points in $C$ are topologically indistinguishable. Commented Aug 2 at 19:34
• Well, then I don't get why this is an example in the Wikipedia page... Commented Aug 2 at 19:43
• You are correct that the example in Wikipedia is invalid. The correct definition for the topology to make this work is given in the answer by Alman. Commented Aug 2 at 21:38
• I have now fixed the incorrect example in en.wikipedia.org/wiki/Regular_space. Commented Aug 3 at 5:20

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 $$K$$.)

• Thank you for your fast and very clear answer. Your proof of why C is not contained in any open supset completes the answer of @Alman. Have a great time! Commented Aug 2 at 20:02

For examples, look at pi-base

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

• WOW! This is a great website! Thank you for your fast and clear answer. As a good example, if you want, you could add the "deleted diameter" topology which I think is very intuitive. Have a great time! Commented Aug 2 at 19:50

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.

• Thank you for your fast and concise answer. This is exactly what I looked for! Have a great time! Commented Aug 2 at 19:53
• You're welcome, sir. I'm glad about that. Notice how, under the same reasoning, you can deduce that, as the usual topology on $\mathbb{R}$ is Functionally Hausdorff, and as the new topology that we have defined is finer than the usual, then it is also Functionally Hausdorff. This is a stronger separation axiom than Hausdorff. I hope you have a great time too. Commented Aug 2 at 19:56
• @K. Jiang's answer is more complete. However I hope time will upvote your Answer to the top as it immediately gives a clear intuition to the answer. Commented Aug 2 at 20:04
• Fair enough. Thank you, sir. Commented Aug 2 at 20:07