Confused as to why the Cofinite Topology is not T2 Separable I think I have a general understanding of the Cofinite Topology over the open interval (0,1), which basically is a collection of open subsets which are determined by individual points 'picked' on the interval.
And if I have an ok understanding of what it means to be T2 Separable, for any two points on the interval, say 'a' and 'b', both of these points need to be able to be placed in their own open sets which are disjoint from each other.  I've seen a counter-example where the interval is divided exactly at 'a' and then exactly at 'b', and then clearly the intersection of the two elements of the Topology is not the null set.
My question is, what if you were to choose an element of the cofinite Topology such that a point in between 'a' and 'b' is picked such that everything to the left of that point was an open interval containing 'a', and then everything to the right of that point was an open interval containing 'b'.  Because the point which was picked has been removed from the interval, the two open sets are clearly disjoint.
Am I misunderstanding something about the definition of T2 Separable or about elements of the cofinite topology?
Thank you in advance.
 A: You’ve misunderstood the cofinite topology. A subset $U$ of $(0,1)$ is open if and only if either $(0,1)\setminus U$ is finite, or $U=\varnothing$. Equivalently, the open sets in this topology are $\varnothing$ and all sets of the form $(0,1)\setminus F$ such that $F$ is a finite subset of $(0,1)$.
Now suppose that $a,b\in(0,1)$, $a\ne b$, $U$ and $V$ are open sets in the cofinite topology such that $a\in U$ and $b\in V$. Clearly $U$ and $V$ aren’t empty, so there are finite subsets $F$ and $G$ of $(0,1)$ such that $U=(0,1)\setminus F$ and $V=(0,1)\setminus G$. Then
$$U\cap V=\big((0,1)\setminus F\big)\cap\big((0,1)\setminus G\big)=(0,1)\setminus(F\cup G)\;.$$
And $F\cup G$ is finite, so it certainly isn’t all of $(0,1)$, and therefore
$$U\cap V=(0,1)\setminus(F\cup G)\ne\varnothing\;,$$
i.e., $U$ and $V$ are not disjoint. This is true no matter what open sets we pick containing $a$ and $b$, so $a$ and $b$ cannot be separated by disjoint open sets, and the space is not Hausdorff.
In fact the same argument shows that in this space no two non-empty open sets are disjoint.
A: If you have open $U$ and $V$ such that $a \in U$, $b \in V$ and $U \cap V = \emptyset$, then this means that $U$ must be cofinite, i.e. $(a,b) \setminus U$ is finite and also $(a,b)\setminus V$ is finite. But then any point $c \in (a,b)$ such that $c \notin ((a,b) \setminus U) \cup ((a,b)\setminus V)$ (there are plenty, as these two sets on the right are finite, and $(a,b)$ is infinite), lies in $U \cap V$ and this contradicts their supposed disjoitness. So $a$ and $b$ cannot be separated by disjoint open sets in the cofinite topology.
