My question comes from reading Pete Clark's reply How do you axiomatize topology via nets?

In the section "Convergence Classes" at the end of Chapter 2 of his book, Kelley lists the following axioms for convergent nets in a topological space $X$

a) If $S$ is a net such that $Sn=s$ for each $n$ [i.e., a constant net], then $S$ converges to $s$.

b) If $S$ converges to $s$, so does each subnet.

c) If $S$ does not converge to $s$, then there is a subnet of $S$, no subnet of which converges to $s$.

d) (Theorem on iterated limits): Let $D$ be a directed set. For each $m\in D$, let $E_m$ be a directed set, let $F$ be the product $D \times \prod_{m \in D} E_m$ and for $(m,f)$ in $F$ let $R(m,f)=(m,f(m))$. If $S(m,n)$ is an element of $X$ for each $m∈D$ and $n\in E_m$ and $\lim_m \lim_n S(m,n)=s$, then $S∘R$ converges to $s$.

He has previously shown that in any topological space, convergence of nets satisfies a) through d). (The first three are easy; part d) is, I believe, an original result of his.) In this section he proves the converse: given a set $X$ and a set $C$ of pairs (net,point) satisfying the four axioms above, there exists a unique topology on $X$ such that a net $S$ converges to $s∈X$ iff $(S,s)∈C$.

The original theorem in Kelley's General Topology says that:

Let $C$ be a convergence class for a set $X$, and for each subset $A$ of $X$ let $A^c$ be the set of all points $s$ such that,for some net $S$ in $A$, $S$ convergences $(C)$ to $s$. Then $c$ is a closure operator, and $(S,s) \in C$ if and only if $8$ converges to $s$ relative to the topology associated with the closure operator $c$.

Kelley didn't explicitly said that the topology that satisfies a net $S \to s$ iff $(S, s) \in C$ is unique, which is not obvious to me either. So I was wondering where the uniqueness comes from? Is it possible that there is other different topology that satisfies a net $S \to s$ iff $(S, s) \in C$, besides the one defined by the closure operator in Kelley's version of the theorem?

If some of the four axioms fail to hold for $C$, is it possible that there is no such topology, and is it possible that there are more than one such topologies?

Thanks and regards!


It is not possible to have two topologies on $X$ with the same nets converging to the same points. To prove it, consider any two distinct topologies $T$ and $T'$ on $X$, and suppose, without loss of generality, that $U$ is an open set in $T'$ but not open in $T$. Let $x$ be a point of $U$ that is not in the $T$-interior of $U$. So you can choose, in each $T$-neighborhood $V$ of $x$, a point $y_V\in V-U$. The collection of $T$-neighborhoods of $x$ is directed by reverse inclusion, and the function assigning to each $V$ in this directed set the chosen point $y_V$ is a net that converges to $x$ with respect to $T$, because, given any $T$-neighborhood $V$ of $x$, we have $y_W\in V$ for all $W\subseteq V$. But this net fails to converge to $x$ with respect to $T'$, because all its points $y_V$ are outside the $T'$-neighborhood $U$ of $x$.

If you consider attempted convergence-classes $C$ for which some of the axioms fail, then there is no topology producing such a $C$ because, as you pointed out earlier in the question, all the axioms are true for convergence in any topology.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.