TITLE: Unique limits of sequences plus what implies Hausdorff?
QUESTION [18 upvotes]: 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?

REPLY [12 votes]: 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.