What tools are used to show a type of convergence is or is not topologizable?

Question

There are many types of convergence. For example,

• in measure theory and probability theory, there are many types of convergence of measurable mappings (random variables).

• in measure theory and probability theory, there are also many types of convergence of measures (over a sigma algebra, a Borel sigma algebra, or ...).

• in topology, there are topology of pointwise convergence and topology of uniform convergence (why?).

For other areas, I believe there exist similar examples. They seem very overwhelming and intimidating (to me at least).

I have seen some discussion on whether there exists a topology or metric or norm such that some type of convergence is consistent wrt them. For example, convergence a.e. is not topologizable, not to mention metrizable or normable.

1. So I was wondering when studying whether a type of convergence is topologizable, metrizable, normable, ..., what tools are are used?

2. Are there some general ways to construct the topologies if they exist?

3. Do all the instances of a certain convergence type form a convergence class? or a convergence space? Do those convergence structures have some general ways to tell if a certain type of convergence is topologizable and to construct such a topology if there is one?

Thanks and regards!

Answer 1

The one that I have seen used: in a topological convergence, suppose we have a sequence $(x_n)$ with the following property, for some fixed point $p \in X$:

(*) If every subsequence $x_{n_k}$ of $(x_n)$ has itself a subsequence that converges to $p$, then $(x_n)$ itself converges to $p$.

(Sketch of proof: suppose $(x_n)$ does not converge to $p$, then there is a neighbourhood of $O$ of $p$ such that no tail of the sequence lies in $O$, or equivalently, for every $k_1$ there exists $k_2 > k_1$ such that $x_{k_2} \notin O$. This allows one to define a subsequence of the original sequence, that lies wholly outside $O$ and so every subsequence of that subsequence also does not converge to $p$. So then the sequence satisfies the negation of the condition (*), so by contrapositive we have shown that (*) holds. )

I believe that in general convergence spaces such convergences (with the condition (*)) are even called topological convergences. [Corrected] The topology of convergence a.e. is in general not a topologisable convergence (it does obey other axioms for a general convergence).

Answer 2

I was going to post this as comment to Henno Brandsma's answer, but it's a little too long for a comment. Nonetheless, this is a follow-up to his answer. Here's an exercise from Tao's "An Epsilon of Room I: Real Analysis" that uses the characterization:

A sequence $x_{n}$ converges to $x$ in a topological space iff every subsequence has a futher subsequence that converges to $x$.

to show that almost everywhere convergence is not, in general, topologisable.

Let $[0,1]$ be equipped with the usual Lebesgue measure. Then the vector space $L^{\infty}([0,1])$ of essentially bounded functions on $[0,1]$ cannot be given a topological vector space structure in which $f_n \to f$ in this topology precisely when it converges almost everywhere.

His hint is to construct a sequence $f_{n} \in L^{\infty}([0,1])$ which does not convergence pointwise almost everywhere to zero, but such that every subsequence has a futher subsequence that converges almost everywhere to zero.

We can let $f_n$ be the so-called "typewriter sequence" (which we can read about here -- the well known sequence that converges in $L^1$ to zero but not pointwise). It is easy to see that this sequence does not convergence pointwise almost everywhere to zero. In fact, it converges pointwise to zero nowhere. How shall we show that every subsequence of $f_n$ has a further subsequence that converges to $0$ almost everywhere? The pigeonhole principle.

Given some subsequence $f_{n_k}$, here is how we will choose our sub-subsequence. At the first step, partition $[0,1]$ into $[0,0.5]$ and $[0.5,1]$. Now, by the pigeonhole principle, infinitely many of these $f_{n,k}$ have support either inside one of these intervals, or have support on both. If it is the latter, then infinitely many functions have support at $0.5$, and then we can just choose a sub-subsequence of such functions whose support shrinks in measure. Thus, let us suppose that infinitely many functions in our subsequence have support in $[0,0.5]$. We can then divide up $[0,0.5]$ into two subintervals and repeat the process. Without loss of generality, all the functions have support on one of these new subintervals, else we can repeat the argument above if infinitely many have support at a point. Iterating this construction, we choose a sub-subsequence of functions which have support in smaller and smaller nested subintervals, and such a sub-subsequence must convergence almost everywhere to zero.