A $T_1$ space which is not Hausdorff

Question

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!

Answer 1

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

Answer 2

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.

Answer 3

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 $X\cap Y\neq\emptyset$.