# A $T_1$ space which is not Hausdorff

More precisely, assume the following definitions.

Definitions. Let $S$ be a topological space.

• $S$ is a $T_1$ space if, whenever $s_1 \neq s_2$ there exists an open set $U_1$ such that $s_1 \in U_1$ but $s_2 \not\in U_1$, and there exists an open set $U_2$ such that $s_2 \in U_2$ but $s_1 \not\in U_2$.
• S is a $T_2$ space or Hausdorff space if, whenever $s_1 \neq s_2$, there exist open sets $U_j$ with $s_j \in U_j$ ($j = 1,2$ ) such that $U_1 \cap U_2 = \varnothing$; that is, $U_1$ and $U_2$ are disjoint.

I'm looking for a $T_1$ space which is not $T_2$. I know that metric spaces are Hausdorff (and even normal), so I discarded them. Moreover, topological spaces with at least two points and trivial topology are not Hausdorff but are not $T_1$ too. Topological spaces of the form $(S, \tau)$, with $S = \{ s_1, s_2 \}$, $\tau = \{ \varnothing, S, \{ s_1 \} \}$ are not $T_1$. (and hence are not Hausdorff.) Nevertheless, I'm sure it can't be a complicated stuff, since this exercise is assigned immediately below the above definitions.

Thanks for help!

• Take an infinite space, say $\mathbb{N}$, with the cofinite topology. Sep 23, 2013 at 18:28
• One thing that is easy to prove is that any finite T$_1$-space is actually discrete (and so is even normal, let alone Hausdorff). So you are in particular looking for an infinite topological space. Sep 23, 2013 at 18:29
• The Zariski topology is another exape of a topology which is $T_1$ but not $T_2$, at least over algebraically closed fields. Sep 23, 2013 at 19:04

Take the natural numbers with the cofinite topology, i.e., $U\subseteq \mathbb N$ is open iff $\mathbb N\setminus U$ is finite or $U=\varnothing$.

• I like this example. Sep 23, 2013 at 18:34
• This is probably the simplest example and at the same time one a beginning student of topology will be very likely to have already seen (which is also a nice feature). Sep 23, 2013 at 18:48
• @ThomasAndrews good catch. I'll edit according. Sep 23, 2013 at 18:50
• Indeed, any point must be closed in a $T_1$ space, and thus all cofinite sets must be open. So the cofinite topology is the smallest $T_1$ topology on any set. Sep 23, 2013 at 19:00
• Thanks, answer and comments are very useful. However, in my book (Singer-Thorpe, Lecture notes on elementary topology and geometry) the cofinite topology is not presented at all. Sep 23, 2013 at 19:15

Let $$X$$ be your favorite infinite set, and let the open subsets of $$X$$ be the empty set and those subsets of $$X$$ with finite complements. This can be shown to be a topology on $$X$$ that is $$T_1$$ but not $$T_2$$.

Incidentally, $$T_1$$ and $$T_2$$ coincide precisely on spaces with finite underlying sets, so there are no non-infinite counterexamples. That is, if we are given a $$T_1$$ topology on a finite set, then it will automatically be a $$T_2$$ topology (the converse holds even on infinite sets), but as we saw above, given any infinite set $$X$$, there is a topology on $$X$$ that is $$T_1$$ but not $$T_2$$. In fact, we can be very specific about the $$T_1$$ topologies on a finite set $$X$$.

Proposition: Suppose $$X$$ is a set. The following are equivalent:

$$(1)$$ $$X$$ is finite.

$$(2)$$ The only $$T_1$$ topology on $$X$$ is the discrete topology (in which every subset of $$X$$ is open).

$$(3)$$ Every $$T_1$$ topology on $$X$$ is $$T_2.$$

The proof of $$(2)\implies(3)$$ is straightforward, and $$(3)\implies(1)$$ can be proved by contrapositive, letting $$X$$ be an arbitrary infinite set and topologizing $$X$$ as described in the first paragraph of my answer.

To prove that $$(1)\implies(2)$$, suppose that $$X$$ is finite with a $$T_1$$ topology. Note/prove that every singleton subset $$\{x\}$$ of $$X$$ is closed (using $$T_1$$), so that every finite subset of $$X$$ is closed (why?), so that every subset of $$X$$ is closed (why?), so that every subset of $$X$$ is open (why?), and so $$X$$ has the discrete topology.

• Could you elaborate on what you mean by that last sentence? Sep 23, 2013 at 18:37
• How is that, Tobias? Sep 23, 2013 at 19:43

Take the space $\mathbb{R}\setminus\{0\}\cup \{a,b\}$ (so remove $0$ and add two other points).

The open sets are those that are open in the usual topology on $\mathbb{R}$ and do not contain $0$ as well as, for each open subset $X\subseteq \mathbb{R}$ with $0\in X$, two open sets, $X\setminus\{0\}\cup\{a\}$ and $X\setminus\{0\}\cup\{b\}$. Also, because we want this to be a topology, we need to add $X\setminus\{0\}\cup \{a,b\}$ for each such $X$, in order to be able to take unions.

Now, it is a nice exercise to check that this space is $T_1$, but that if $X$ and $Y$ are open sets such that $a\in X$ and $b\in Y$ then $X\cap Y\neq\emptyset$.

• Thank you, your answer seems to be the simplest one. However, I'm not sure about why we do need to remove $0$. Sep 23, 2013 at 19:38
• @Federico We don't really. We could also just add one point and add those open sets containing it. The reason I added two points was to emphasize the symmetry between them (the two newly added points behave basically the same). Sep 23, 2013 at 19:39