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You can set some kind of convergence in a space of functions without using some metric or topology or sigma field?

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So, convergence-a.e. does not fit because it goes via the definition of a $\sigma$-field? – Ilya Mar 29 '12 at 15:02
a.e. convergence involves a topology or a metric or the like on the codomain. But "a.e." is about the domain. – Michael Hardy Mar 29 '12 at 15:09
@Ilya I would suppose so, since for convergence a.e, you need some notion of convergence already, which you would then relax over the measure zero sets.. – Galois Group Mar 29 '12 at 15:09
The convergence in metric depends on a collection of subsets. Namely the topology generated by the metric. The a.e. convergence depends of a collection of subsets ( and the measure too). Namely the sigma field. It seems to me that the notion of convergence depends on a collection of sets convenient. My question would be in that direction. Not specific to my question does not limit the possibilities. – MathOverview Mar 29 '12 at 15:28
@Michael: I didn't get your comment, sorry. A.e. is about a domain - but convergence a.e. does not involve any topology on the codomain, does it? – Ilya Mar 29 '12 at 15:43
up vote 2 down vote accepted

Here is a trifle of an example that seems to suffice your requirement.

Let $A$ and $B$ be any sets, and consider the space $B^A$ of functions from $A$ to $B$. Also, let $\omega$ be a nonprincipal ultrafilter on $\mathbb{N}$. That is, $\omega$ is a maximal filter on $\mathbb{N}$ that contains no finite sets. (The existence of such filter is ensured by the Axiom of Choice.)

Then for a sequence $(f_n) \subset B^A$ of functions and a function $f \in B^A$, we say $$ f_n \stackrel{\omega}{\longrightarrow}f$$ if for every $x \in A$, the set $\{ n \in \mathbb{N} : f_n (x) = f(x) \}$ is contained in $\omega$.

It is easy to prove the uniqueness of the limit, and if there is an algebraic structure on $B$, it easily follows that this notion of limit is compatible with the operations on $B$.

But it does not capture any useful concept of 'closeness' (rather, it is just a description incognito of 'equal a.e. $n$ pointwise'), so it seems of little importance to consider this kind of notion.

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Trifle perhaps, but also large research area. – André Nicolas Mar 29 '12 at 17:51
It’s arguably still topological: it’s just topological $\omega$-convergence when $B$ is given the discrete topology. – Brian M. Scott Mar 29 '12 at 17:57
@AndréNicolas: Im sorry if my answer seemed to disrespect a certain branch of mathematics. I'm not intended to disparage the ultrafilter concept. Rather, I just wanted to figure out that the particulal instance described in the body text is trivial. – Sangchul Lee Mar 29 '12 at 21:42
@BrianM.Scott : And it also conceals nearly a concept of measure since ultrafilter is a special cond of finite additive measure on the power set of natural numbers. What I tried is to give an example which is deprived of many interesting structures. – Sangchul Lee Mar 29 '12 at 21:45
I think that you probably did remove about as much topology and measure as can be removed. – Brian M. Scott Mar 29 '12 at 21:49

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