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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!

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up vote 4 down vote accepted

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.

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