Why the separation axiom $T_0$ isn't implied by the others? I was reading here 
https://en.wikipedia.org/wiki/Separation_axiom#Relationships_between_the_axioms
and it gives two different kind of axioms, one that has $T_0$ property and one that hasn't. My definition of $T_0$ space is that two distinct points are distinguishable, so that there is a neighborhood of one which doesn't contain the other. But I thought $T_0$ was implied by pretty much everything else. Why isn't that so?
 A: You are right that, say, a Hausdorff space, that is $T_2$ holds, also satisfies $T_0$ and thus "non-$T_0$ version of a $T_2$ space" looks contradictory.  
However, what is meant by, e.g., "non-$T_0$ version of a $T_2$ space"  is not a space that is "$T_2$ but not $T_0$."
What is meant is a space that is not (necessarily) $T_0$ but its Kolomogorov quotient is $T_2$. 
Informally, this Kolomogorov quotient is the space one obtains by removing the failure of $T_0$ by declaring points that have the same neighborhoods to be equal.  
A: The non-$T_0$ version of Hausdorff is called $R_1$ and it is defined as: 

For all $x,y$ that are topologically distinct, there are open neighbourhoods $U$ of $x$ and $V$ of $y$ such that $U \cap V = \emptyset$.

Where $x$ and $y$ topologically distinct means that there is an open set that contains $x$ but not $y$ or an open set that contains $y$ and not $x$.
Formulated from the negative side: two points $x,y$ are topologically indistinguishable iff for all open sets $O$ of $X$: $x \in O \leftrightarrow y \in O$. (And then $x$ and $y$ are topologically distinct iff they're not topologically indistinguishable). For example in an indicrete space all points are topologically indistinguishable as the only non-empty open set is $X$.
SO an indiscrete space $X$ is always $R_1$, trivially, as there are no topologically distinct points to separate.
Another example, take $\mathbb{Z}$ with the topology generated by the partition $\{\{2n, 2n+1\}; n \in \mathbb{Z} \}$. Then $0$ and $1$ are topologicaly indistinguishable, as are all pairs $2n$, $2n+1$. But e.g. $n,m$ that are not of this form are topologically  distinct, and also have disjoint neighbourhoods. So this space is $R_1$ but not $T_0$ and certainly not $T_2$. It's like we have "clots" of points that stick together, but different such "clots" can be separated from each other.
