I am working with the topology $T=\{\emptyset,\{1\},X\}$ where $X$ is the set $\{1,2,3\}$
I need to prove the the topology $T$ is not Hausdorff, I was going to use the definition of a Hausdorff Space so though the most logical way would be to prove the negation i.e. $$\text{there exists} x,y \in X \text{ with } x\ne y, \text{for every open set } U,V \text{ in }X \text{ with } x\in U, y\in V \text{ and } U\cap V\ne \emptyset$$
But this is where I'm having trouble, I know I need to take the elements of X and put them into this, but in terms of getting it written out I'm having trouble in terms of what comes next
I'm assuming my $x$ or $y$ value would have to be $1$ as this means any intersection would not be empty?
Any ideas or help would be great
The only open subset of $X$ to which $2$ belongs is $X$ itself. The same thing holds for $3$. So, there are no disjoint open subsets $U$ and $V$ of $X$ such that $2\in U$, $3\in V$, and $U\cap V=\emptyset$. Actually, this not only shows that $(X,T)$ is not Hausdorff, as it actually shows that it's not a $T_0$ space.