Unique limits of sequences plus what implies Hausdorff? It is known that there are non-Hausdorff spaces which admit unique limits for all convergent sequence (see here) and it is also known that unique limits for nets implies Hausdorff.
What I am wondering is, if there is a (somehow weak) condition which one should add to "unique limits of sequences" to obtain a Hausdorff space. Would, for example, some countability help?
Somehow in the same direction: What is the central property which is needed for a space such that it can be non-Hausdorff but has unique sequence limits? Is there a whole class of non-Hausdorff spaces which admit unique limits for convergent sequence?
 A: Here's an example of a space which is not Hausdorff but which has unique limits...
Let $X = \mathbb{R}$ with the cocountable topology, i.e. a set is open iff its complement is countable. Clearly any two open sets intersect, because $\mathbb{R}$ is uncountable. So $X$ is non-Hausdorff. Now, suppose $(x_n)$ is a sequence which converges to $x$. Then $C =$ {$x_n\;|\;x_n\neq x$} is closed because it's countable. So $X-C$ is a neighborhood of $x$ and this means there is some $N$ such that for all $n>N$ $x_n\in X-C$, i.e. $x_n=x$ for large $n$. This means if $x_n\rightarrow y$ then $y=x$, proving limits are unique.
A: This past weekend, entirely by chance, I came across a published paper that used the term "US-space" for the class of topological spaces having the property that no sequence can converge to more than one point. The google search just below seems to bring up some things that might be of use to you:
http://www.google.com/search?q=%22US-space%22+convergence+sequence
A: First countable is enough. Let $x\neq y$ be two points in your space that cannot be separated by neighborhoods. Let $O_1,O_2,\ldots$ form a neighborhood base of $x$ and let $U_1,U_2,\ldots$ form a neighborhood base for $y$. Choose a sequence $(z_n)$ such that $z_n\in O_n\cap U_n$ for all $n$. Now $(z_n)$ converges to both $x$ and $y$.
A: Here is an answer to Dirk's last question, "Is there a class of non_Hausdorff spaces in which convergent sequences have unique limits?"
Yes. The so called KC-spaces or  maximal compact spaces. These are spaces such that every compact subspace is closed. 
(The 1967 Monthly article of Wilansky, "Between $\mathrm T_1$ and $\mathrm T_2$" (MSN), subsumes, references, or implies all of the following.)
In a KC-space, convergent sequences have unique limits.
(Suppose $x_n\to x$ in the KC space $X$. The set $\{x,x_1,x_2,\dotsc\}$ is compact and hence closed. Thus, if $y$ is not in the set $\{x,x_1,x_2,\dotsc\}$, then the open set $X \setminus \{x,x_1,x_2,\dotsc\}$ shows it is false that $x_n\to y$. Thus, if $x_n\to y$, then $y=x$ or $y=x_n$ for some $n$. If $y=x_n$ for infinitely many indices $n$ then $y=x$ (since every KC space is $\mathrm T_1$ (since singletons are compact) and since constant sequences have unique limits in a $\mathrm T_1$ space). If $y=x_n$ for finitely many indices $n$ then (deleting $y$ from the sequence $x_1,x_2,\dotsc$) we are left with a subsequence $z_n\to x$, the knowledge that $y$ is not equal to any $z_n$, and the knowledge that $y$ is in the set $\{x,z_1,z_2,\dotsc\}$, and we conclude that $y=x$).
To exhibit a large class of non-Hausdorff KC spaces, let $X$ be a non-locally-compact metric space (for example, the rationals) and let $Y=X \cup \{y\}$ denote the Alexandroff compactification of $X$ (i.e., $V$ is open in $Y$ if $V$ is open in $X$ or if $Y\setminus V$ is a compact subspace of $X$).
The space $Y$ is a KC space, but $Y$ is not Hausdorff.
