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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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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0I think that you probably did remove about as much topology and measure as can be removed. – 2012-03-29