A sequence in a Hausdorff space and in a space that is not Hausdorff. 
Let $X$ be a topological space and $\{x_n\}_{n=1}^{\infty}$ a sequence in $X$.
Show that if $X$ is Hausdorff, $x_n \rightarrow x \:$, $x_n \rightarrow y \:$  implies $x=y$.
Give an example that shows that if $X$ is not Hausdorff then this is not necessarily true.

A sequence is convergent, $x_n \rightarrow x \:$, if there is an element $x \in X$ such that for every open neighborhood $U$ of $x$ there exists a $n_0$ such that $n > n_0$ implies $x_n \in U$.
If $X$ is Hausdorff and $x$ and $y$ are different there exists open disjoint sets $x \in U$ and $y \in V$.
Because all but a finite number of $x_n$ lies in $U$ according to the definition of that $x_n \rightarrow x \:$ then only a finite number of $x_n$ can be in $V$, which is a contradiction to that $x_n \rightarrow y \:$.
But i have a hard time to come up with an example that shows that this not necessarily is true when $X$ isnt Hausdorff.
Can anyone give me a hint?
Thanks.
 A: Take the antidiscrete space with more than one point. Every sequence converges to any point in the space.
Antidiscrete space $X$ has just 2 open sets: empty set and itself.
Let $a_n$ be a sequence in $X$.
The claim: the sequence converges to $x$ for any $x\in X$.
We need to show that any open set containing $x$ contains all but finite elements of the sequence. But there is only one open set containing $x$, which is the set $X$ itself. And we know that $X$ contains all elements of $a_n$, basically because $a_n$ is the sequence in $X$.
A: Take $\mathbb{N}$ in the cofinite topology (the only closed sets are the finite ones (including the empty set) and $\mathbb{N}$ itself). Take $a_n$ to be any sequence where all values are different, like $a_n = n$ or $a_n = 2n$ etc.. Then $(a_n)$ converges to every point $m$ of $\mathbb{N}$, because the only open sets that contain $m$ are of the form $O = \mathbb{N} \setminus F$, where $F \subset \mathbb{N}$ is finite. But after some initial segment the values of $a_n$ are never in $F$ (as $F$ is only finite), so all values from some value onwards are in $O$. As this holds for all $O$, $(a_n) \rightarrow m$ for all $m$.
So any two of these sequences have all points of the space as their limit as well and so the limits are very non-unique. The cofinite topology is $T_1$, so one separation axiom below Hausdorff, making it a "sharper" example than the indiscrete topology. 
A: I'll present a fairly general way of coming up with examples of sequences which converge to more than one point. Recall that a space $X$ is called T1 (or Fréchet) if given any two distinct points $x,y \in X$ there is an open set which contains $x$ but not $y$.
Suppose that $X$ is not T1.  (Examples of such spaces can be found in π-Base with this search.)  Then there are distinct points $x,y \in X$ such that every open neighborhood of $x$ contains $y$. Consider the constant sequence $( x_n )_n$ where $x_n = y$ for each $n$. Note that clearly this sequence converges to $y$, but by the assumption on the chosen points we also have that the sequence converges to $x$.
One can also show that if a space has the property that all sequences converge to at most one point, then the space must be T1.

As a final aside of sorts, note that there are non-Hausdorff spaces in which every sequence converges to at most one point. An example would be an uncountable set given the co-countable topology, wherein the only convergent sequences are the eventually constant sequences, which converge only to their eventually constant value.
A: This should work also:
Let $X = \mathbb{R}$ with the topology $\{(-\infty, x): x \in [-\infty, \infty]\}$
$x_n \rightarrow x \:\: \Rightarrow \:\:x_n \rightarrow y \:\:\: \forall\:\: y \geq x$
