I am reading through Munkres pg 99 and it says without proof, "the real line in the finite complement topology is nota hausdorff space" i am having trouble getting started with proving this claim .. thank you for a push in the right direction.
I think this would be proved by saying along the lines that (X- $U_{2}\bigcup U_{1})$ = (X - $U_{1}$)$\bigcap (X - U_{2}) \neq \emptyset$ then not hausdorff. ?
Suppose that $X$ is equipped with the cofinite topology and that $x,y$ are distinct elements.
If it is Hausdorff then open sets $U,V$ must exist with $x\in U$, $y\in V$ and $U\cap V=\emptyset$.
Then $U\subseteq V^c$ tells us that $U$ is finite and consequently $X=U\cup U^c$ is finite.
The real line however is not finite, so...
This is not true for $\mathbb{R}$ or any other infinite space $X$ with the finite complement (cofinite) topology.
Normally, if you wanted to prove $X$ is not Hausdorff, you would just need to prove that there is one specific pair of $p_1$ and $p_2$ where the two points cannot be separated by disjoint open sets around each one. However, in this case if $X$ is infinite, you can't do this for any $p_1$ and $p_2$ pair.
If you had $p_1$ and $p_2$ points in $X$ and $U$ any neighborhood of $p_1$, there is no neighborhood of $p_2$ in $X\ \backslash\ U$ because $X\ \backslash\ U$ is finite (and so could not contain any cofinite neighborhood for us to put around $p_2$).
The $\backslash$ symbol for me means complement, i.e. $X\ \backslash\ U=X\cap U^c$ which (since $X$ is the whole space) is $U^c$.
In fact, ignoring $p_1$ and $p_2$, it is impossible to have two two disjoint nonempty open sets $U_1$ and $U_2$ (by virtue of being open and disjoint, each one seems to insist that the other is finite).
This cofinite topology is nice, as an example, because it is a Fréchet space (aka T1, where Hausdorff is T2). To be T1, you must satisfy the weaker condition: For every $p_1$ and $p_2$ in $X$, there is a neighborhood of $p_1$ not containing $p_2$ and a neighborhood of $p_2$ not containing $p_1$.
Notice that the difference between this and Hausdorff is that $U_1$ and $U_2$ don't have to be disjoint -- so long as they don't contain both points.
Given points $p_1$ and $p_2$, you can take $U_1$ to be $X\ \backslash\ \{p_2\}$ and likewise for $U_2$. Clearly $U_1$ is cofinite and doesn't contain $p_2$ (its complement is just $\{p_2\}$ after all!). Likewise for $U_2$.
But of course, $U_1$ and $U_2$ have significant overlap!
Suppose, if possible, that $R$ is a Hausdorff space, then for any two distinct points $x_1$ and $ x_2$, there exist open sets $ U$ and $V$ such that $x_1∈U,\; x_2∈V$ and $U∩V=∅$.
Then since $U∈τ,V∈τ$ and $ τ$ is a finite complement topology, $U^c$ and $V^c$ are finite.
We know that, $U⊂V^c$; hence U is finite too. Then, $R=U∪U^c$ is finite. It is a contradiction. Therefore, it follows that the real line $R$ in the finite complement topology is not a Hausdorff space.
