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.

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2 Responses to Student’s Question: Unique Limits of Convergent Nets and Separation Axioms

  1. Felix V. says:

    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 $y$.

    • Matt R. says:

      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!

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