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.
Best Answer
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 $X$.