## Student’s Question: Unique Limits of Convergent Nets and Separation Axioms

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. $(X,\tau)$ is a Hausdorff space if and only if every net has at most one limit.

Proof. I leave the $\Rightarrow$ direction as an exercise. Suppose $(X,\tau)$ is not Hausdorff, and let $x,y\in X$ be distinct points with no disjoint neighborhoods. Let $\mathcal{N}_{x}$ and $\mathcal{N}_{y}$ respectively denote the collections of open neighborhoods of $x$ and $y$. Partially order the product $\mathcal{N}_{x}\times\mathcal{N}_{y}$ by $U_{\beta}\times V_{\beta}\geq U_{\alpha}\times V_{\alpha}$ $\Leftrightarrow U_{\beta}\subset U_{\alpha}$ and $V_{\beta}\subset V_{\alpha}$ to obtain a directed set. By assumption, for every $U_{\alpha}\times V_{\alpha}\in\mathcal{N}_{x}\times\mathcal{N}_{y}$, we can choose $z_{\alpha}\in U_{\alpha} \times V_{\alpha}$ to obtain a net $(z_{\alpha})$.

I claim that $z_{\alpha}$ converges to both $x$ and $y$. Indeed, let $U$ and $V$ be open nbhds of $x$ and $y$, respectively. For every $U_{\alpha}\times V_{\alpha}\geq U\times V$, $z_{\alpha}\in U\cap V$. $\Box$

Today, a student asked me whether a topological space which has the unique sequential limit property is necessarily $T_{1}$; i.e., singletons are closed. I mistakenly said that I believed the unique sequential limit property was equivalent to the $T_{1}$ separation axiom. I’d now like to show that the unique limit property is, in fact, stronger than $T_{1}$.

Proposition. If $(X,\tau)$ is a topological space in which convergent sequences have unique limits, then $X$ is $T_{1}$.

Proof. If $X$ has unique limits, then the constant sequence $x_{n}:=x$, for all $n$, uniquely converges to $x$. Whence, for any $y\neq x$, there exists an open neighborhood $U\ni y$ such that $x\notin U$, which implies that $X\setminus\left\{x\right\}$ is open. $\Box$

We now prove that the converse to the proposition is false. Consider $\mathbb{Z}$ equipped with the cofinite topology: a subset $U\subset \mathbb{Z}$ is open if and only if $U=\emptyset$ or $U^{c}$ is finite. I claim that this space is $T_{1}$. Indeed, let $x$ and $y$ be distinct integers. Define sets $U_{x}:=\left\{n\in\mathbb{Z}: n \neq y\right\}$ and $U_{y}:=\left\{n\in\mathbb{Z}:n\neq x\right\}$. Clearly, $x\in U_{x}$ and $y\in U_{y}$, and $U_{x}, U_{y}$ are open since $U_{x}^{c}=\left\{y\right\}$ and $U_{y}^{c}=\left\{x\right\}$.

Consider the sequence $(x_{k})_{k=1}^{\infty}$, where $x_{k}:=k$. I claim that every $x\in\mathbb{Z}$ is a limit of this sequence. Indeed, let $U$ be an open neighborhood whose complement consists of the distinct points $y_{1},\cdots,y_{m}$. Set $N:=\max\left\{y_{1},\cdots,y_{m}\right\}$. For all natural numbers $k\geq N$, $k\in U\Longleftrightarrow x_{k}\in U$. Thus, we have exhibited a sequence in this $T_{1}$ 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.

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 $y$.