You can set some kind of convergence in a space of functions without using some metric or topology or sigma field?
Convergence without metric or topology or sigma field.
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0So, convergence-a.e. does not fit because it goes via the definition of a $\sigma$-field? – 2012-03-29
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0a.e. convergence involves a topology or a metric or the like on the codomain. But "a.e." is about the domain. – 2012-03-29
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1@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.. – 2012-03-29
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0The 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. – 2012-03-29
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0@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? – 2012-03-29
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0The meaning of the statement $\lim\limits_n f_n(x) = f(x)$ for _almost all_ $x$ in the domain depends on the meaning of the statement that $\lim\limits_n f_n(x) = f(x)$ for some particular $x$ in the domain. The terms $f_n(x)$ belong to the codomain. When you talk about a sequence of points in the codomain converging to a point in the codomain, that involves a topology or metric or the like in the codomain. – 2012-03-29
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0So how do we talk about convergence - well we have an indexed sequence of points which get "close" to a limit point, and which get (in some general sense) "closer" as the index increases that they become (and this is a expression to avoid "limit") "so close as to be indistinguishable" from it. Now there is probably a way of doing this with specific sequences and limits which avoids something like topology - but to work for all points of the space and all sequences of interest, you have to capture a general concept - as topology does. – 2012-03-29
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0It seems to me that the notion of Convergence requires some notion of closeness of points, which I cannot think of a way to do that without a collection of privileged sets to work with. But, I am no expert. – 2012-03-29
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0If it has any relevance: There is the notion of *convergence spaces*, which generalize topological spaces. – 2012-03-29
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0Doesn't pointwise convergence fit this definition? – 2012-03-29
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0@Michael: ah, sure - I confused the codomain with the space of functions itself – 2012-03-29
1 Answers
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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0Trifle perhaps, but also large research area. – 2012-03-29
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0It’s arguably still topological: it’s just topological $\omega$-convergence when $B$ is given the discrete topology. – 2012-03-29
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0@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. – 2012-03-29
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0@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. – 2012-03-29
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0I think that you probably did remove about as much topology and measure as can be removed. – 2012-03-29