Earlier this week in lecture, my class went over the proof that convergent nets in a Hausdorff space necessarily have unique limits. Actually, the majority of my students don’t know what nets are, so we proved this result in the special case of sequences; however, the argument is the same. After finishing the proof, I told my students that the converse is not true: there exists a non-Hausdorff space in which convergent sequences have unique limits. I did not give them an example; I did them one better by asking them to find such a space on their next problem set! It is true, however, a space in which every convergent net has a unique limit is Hausdorff.

Proposition.is a Hausdorff space if and only if every net has at most one limit.

Proof. I leave the direction as an exercise. Suppose is not Hausdorff, and let be distinct points with no disjoint neighborhoods. Let and respectively denote the collections of open neighborhoods of and . Partially order the product by and to obtain a directed set. By assumption, for every , we can choose to obtain a net .

I claim that converges to both and . Indeed, let and be open nbhds of and , respectively. For every , .

Today, a student asked me whether a topological space which has the unique sequential limit property is necessarily ; i.e., singletons are closed. I mistakenly said that I believed the unique sequential limit property was equivalent to the separation axiom. I’d now like to show that the unique limit property is, in fact, stronger than .

Proposition.If is a topological space in which convergent sequences have unique limits, then is .

*Proof. *If has unique limits, then the constant sequence , for all , uniquely converges to . Whence, for any , there exists an open neighborhood such that , which implies that is open.

We now prove that the converse to the proposition is false. Consider equipped with the cofinite topology: a subset is open if and only if or is finite. I claim that this space is . Indeed, let and be distinct integers. Define sets and . Clearly, and , and are open since and .

Consider the sequence , where . I claim that every is a limit of this sequence. Indeed, let be an open neighborhood whose complement consists of the distinct points . Set . For all natural numbers , . Thus, we have exhibited a sequence in this topological space which does not have a unique limit.

There is a class of topological spaces called *US spaces *defined by the property that convergent sequences have unique limits. As we have shown, it lies strictly between the class of Hausdorff spaces and the class of T1 spaces.

Exercise 31 in §4.3 of Folland’s “Real Analysis” book claims that a space is Hausdorff iff every net in X converges to at most one point. This seems to contradict the main statement in the introduction.

The proof idea given is to consider 4,y which do not have disjoint neighborhoods and then to consider the directed set $N_x \times N_y$, where $N_x, N_y$ are the families of neigbhorhoods of $x$/$y$. If we then choose some $z_{U,V} \in U\cap V$ for each $(U,V) \in N_x \times N_y$, we get a net which converges to $x$ as well as to $y$.

You’re right. It should read that there are non-Hausdorff spaces such that every convergent sequence has a unique limit. I’ll correct it. Good catch!