# 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?

• The link in your question doesn't work for me. Anyway, it's easy to produce non-Hausdorff spaces with unique limits. The Frechet Topology has unique limits and is anti-Hausdorff (all open sets intersect). Another example is the co-countable topology on $\mathbb{R}$. See en.wikipedia.org/wiki/Fr%C3%A9chet_space and en.wikipedia.org/wiki/Cocountable_topology – David White Sep 7 '11 at 12:45
• I believe having unique limits implies the space is $T_1$, so perhaps the question boils down to $T_1$ plus what implies $T_2$. – David White Sep 7 '11 at 12:50
• I'm a bit confused. The Wikipedia page says that Frechet spaces are indeed Hausdorff. Also: Why do you have unique limits of sequences in the cocountable topology? – Dirk Sep 7 '11 at 13:46
• It seems you're right about Frechet spaces. I admit that I was just quoting a line from a topology course I took some time ago, and I didn't stop to question whether or not it's true. However, the cocountable one is right and I'll post the details as an answer – David White Sep 7 '11 at 15:23
• Looking back at my notes, it wasn't Frechet spaces, but rather the Frechet topology on any space $X$, where you define $A$ to be closed iff $A$ is the set of limits of sequences in $A$. This topology can be used to construct an example of an anti-Hausdorff space with unique limits, but the answer below is just as good. – David White Sep 7 '11 at 15:31

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 $y$.

• Thanks! To clarify: First countable + non-Hausdorff implies "non-unique limits" and hence, "unique limits" + first countable implies Hausdorff, right? – Dirk Sep 7 '11 at 10:45
• First countable is a nice answer. I don't think you're going to do better because you kind of need at least this assumption to say anything about sequences – David White Sep 7 '11 at 12:50
• Yes as David said there's a general rule of thumb that as soon as your space is first countable all topological informations are encoded in the behavior of sequences while in non first countable spaces sequences are not sufficient anymore to describe your topology. So you can take this as a general instruction without worrying anymore ...btw can someone proof that information necessarily will be lost by doing so? That could be sth like there is certainly some subset whos closure is bigger than the set of limits of sequences from the set – C-Star-W-Star Apr 9 '14 at 15:10
• I opened a new thread on this: math.stackexchange.com/questions/746828/… – C-Star-W-Star Apr 9 '14 at 16:01

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.

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.

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: