Sequences in Non-Hausdorff Spaces I was told that in any space that is not Hausdorff there are at least two points such that any sequence converges to one iff it converges to the other.  I don't know how to prove this.  Could I have any help?
 A: The statement is false even for $T_1$ spaces. Let $X$ be an uncountable set, and give $X$ the co-countable topology: $V\subseteq X$ is open iff either $X\setminus V$ is countable, or $V=\varnothing$. Then all countable subsets of $X$ are closed, to all finite subsets of $X$ are closed, and $X$ is therefore $T_1$. On the other hand, a sequence $\langle x_n:n\in\Bbb N\rangle$ is convergent iff it is eventually constant, i.e., iff there are an $x\in X$ and an $n_0\in\Bbb N$ such that $x_n=x$ for all $n\ge n_0$, and in that case it converges to $x$ and only to $x$.
A: This is false, I think.
Let $X = \{a,b\}$ and define a topology on $X$ by delaring $U$ open iff $a\in U$ or $U = \emptyset$.
Then $X$ is not Hausdorff as $a$ and $b$ are two points which can not be separated by disjoint open sets (since every open set containing $b$ must contain $a$).
Then the constant sequence $b,b,b,b,...$ converges to $b$ but not $a$.  To see this, notice that the sequence is not eventually in the open set $\{a\}$.
A: The statement "there are at least two points such that any sequence converges to one iff it converges to the other" is somewhat ambiguous.
It could be read as "there are $x, y$ with $x \ne y$ such that any sequence that
converges to $x$ also converges to $y$". In this case it is equivalent to "there are $x, y$ with $x \ne y$ such that every neighbourhood of $y$ is also a neighbourhood of $x$", which is the negation of T1.
Alternatively, it could mean "there are $x, y$ with $x \ne y$ such that any sequence that converges to $x$ also converges to $y$ and any sequence that
converges to $y$ also converges to $x$". In that case it is equivalent to
"there are $x, y$ with $x \ne y$ such that every neighbourhood of $y$ is also a neighbourhood of $x$ and every neighbourhood of $x$ is also a neighbourhood
of $y$", which is the negation of T0.
Either way, T2 does not come into it.

EDIT:
Reading the question again, I noticed that I overlooked the second f in "iff".
Taking that into account, the second interpretation is the right one.
