1

I was wondering if there exists a topology $\tau$ in $L^1(\Omega,\mathcal{L}^d), \Omega\subset \mathbb{R}^d$ open bounded, such that a sequence $v_n\xrightarrow{b} v$ if and only if $v_n \xrightarrow{\tau} v$.

For $v_n \xrightarrow{b} v$ I mean $v_n$ converges in the sense of biting to $v$, i.e. there are $E_i\subset \Omega$ countably many measurable sets, $\mathcal{L}^n(\Omega\setminus E_i)\downarrow 0$ and $v_n$ converges weakly to $v$ in $L^1(E_i,\mathcal{L}^d)$ for every $i\in \mathbb{N}$.

Tommaso Seneci
  • 1,499
  • Define a set $K$ to be closed if for every ${v_n} \subset K$ such that $v_n\xrightarrow{b} v$, $v$ is also in $K$. This may or may not give you exactly the topology you are after (it's been a long time since I've looked into it, but there is a reason we generally don't define topologies by convergence of sequences), but it will be close. – Paul Sinclair Mar 06 '20 at 04:19
  • I doubt this is going to work, as in general defining closed sets via that method above does not give rise to topologies. Think of pointwise Lebesuge almost everywhere converges for instance. – Tommaso Seneci Mar 06 '20 at 14:11
  • Let's see: Are the $\emptyset$ and the whole space both "closed"? yes. Is the union of two "closed" sets also "closed"? yes. Is the intersection of any collection of "closed" sets also "closed"? yes. That means my definition of closed does indeed define a topology $\tau$. Certain topologies cannot be defined by sequences (e.g., the closed long line), but a concept of sequential convergence can be used to define topologies, if convergence $\implies$ subsequence convergence. The only question is whether or not $v_n \xrightarrow {\tau} v \iff v_n \xrightarrow {b} v$. – Paul Sinclair Mar 06 '20 at 17:14
  • So your claim is that every notion of sequential convergence defines a topology? What about Lebesgue-almost everywhere convergence? Notice that you can definitely always define a topology that way, but there is no guarantee that convergence sequences will be equivalent. Read this for instance https://math.stackexchange.com/questions/2493715/topology-induced-by-a-convergence-notion – Tommaso Seneci Mar 07 '20 at 19:39
  • Gee, thank you for informing me of what I very clearly said in both of my previous comments (which is, by the way, the reason I made comments instead of an answer). And yes, any concept of sequential convergence that implies convergence of subsequences to the same limit (necessary to show the union of two closed sets is closed) will define a topology. – Paul Sinclair Mar 07 '20 at 19:52

0 Answers0