# Any relations between the weak topology on a Banach Space and the weak topology on CW complexes?

I'm learning about CW complexes, which we'll say are topological spaces $X$ that admit a filtration $\emptyset \subset X^0 \subset X^1 \subset \cdots \subset X^n \subset \cdots$, with $X=\bigcup X^n$, plus some other requirements. They are endowed with the weak topology. That is, a subset $A$ is closed in $X$ iff $A\cap X^n$ is closed for each $X^n$.

In functional analysis, I learned that the weak topology on a Banach space $Y$ is the coarsest topology making the elements in the dual space $Y^*$ continuous.

Do these two topologies share anything more than a name? Why is it they are both given the label 'weak'?

• – Paul Frost Jul 23 '19 at 13:29

The topology on a CW complex $X$ is the final topology on $X$ with respect to the inclusions $X^n \to X$, ie. it's the finest topology on $X$ such that these inclusions are continuous.

On the other hand, as you say the weak topology is the initial topology wrt continuous linear functionals. These two notions are dual (initial and final topologies).

I doubt there's more of a connection than that. These are two different fields of mathematics (algebraic topology vs "analysis").

• Ah, I see what the relation is now. These topologies are the coursest topologies that preserve continuity of the maps that matter in each context. (btw...I do think you mean coursest and not finest, as the discrete topology also gives this and is much finer than the weak topology on CW complexes.) – Rachel Sep 18 '14 at 12:50
• No, no, I do mean finest. The discrete topology on $X$ doesn't make the inclusions $X^n \to X$ continuous in general; if $X$ has cells of dimension $n > 0$, the preimage of a point belonging to the interior of an $n$-cell in the $n$-skeleton will not be open. – Najib Idrissi Sep 18 '14 at 13:10
• On the other hand the coarsest topology on $X$ that makes all the inclusions continuous is the indiscrete topology. Notice that the maps go into $X$, not from $X$ like in the case of the linear functionals. – Najib Idrissi Sep 18 '14 at 13:11