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!

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    $\begingroup$ Take an infinite space, say $\mathbb{N}$, with the cofinite topology. $\endgroup$ – Daniel Fischer Sep 23 '13 at 18:28
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    $\begingroup$ 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. $\endgroup$ – user642796 Sep 23 '13 at 18:29
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    $\begingroup$ The Zariski topology is another exape of a topology which is $T_1$ but not $T_2$, at least over algebraically closed fields. $\endgroup$ – Oliver Braun Sep 23 '13 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 $U=\varnothing$.

  • $\begingroup$ I like this example. $\endgroup$ – Rustyn Sep 23 '13 at 18:34
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    $\begingroup$ 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). $\endgroup$ – Tobias Kildetoft Sep 23 '13 at 18:48
  • $\begingroup$ @ThomasAndrews good catch. I'll edit according. $\endgroup$ – azarel Sep 23 '13 at 18:50
  • $\begingroup$ 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. $\endgroup$ – Thomas Andrews Sep 23 '13 at 19:00
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    $\begingroup$ 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. $\endgroup$ – Federico Sep 23 '13 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.

  • $\begingroup$ Could you elaborate on what you mean by that last sentence? $\endgroup$ – Tobias Kildetoft Sep 23 '13 at 18:37
  • $\begingroup$ How is that, Tobias? $\endgroup$ – Cameron Buie Sep 23 '13 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 $X\cap Y\neq\emptyset$.

  • $\begingroup$ Thank you, your answer seems to be the simplest one. However, I'm not sure about why we do need to remove $0$. $\endgroup$ – Federico Sep 23 '13 at 19:38
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    $\begingroup$ @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). $\endgroup$ – Tobias Kildetoft Sep 23 '13 at 19:39

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