# What do the open sets in the weak* topology look like?

I was studying functional analysis a few months ago and asked a similar question (On the definition of weak and weak-* topologies). I thought I had understood weak* topologies rather well, until I started thinking about them some more today. I realized I am unable to even "visualize" the open sets.

Suppose $$X$$ is a Banach space over a field $$\mathbb{F}$$ and let $$X^*$$ be its dual. The weak topology is by definition the weakest topology making all $$f \in X^*$$ continuous. From this definition it is easy to visualize the open sets in the weak topology on $$X$$; simply take any open $$U \subset \mathbb{F}$$ then we define $$f^{-1}(U)$$ to be open in the weak topology for any $$f \in X^*$$.

I am having trouble carrying out a similar picture for the weak* topology. For example, let $$X^{**}$$ denote the double dual of $$X$$ and again let $$U \subset \mathbb{F}$$ be open. Consider all linear functionals of the form $$\Lambda_x(f) = f(x)$$. Then what do the open sets in the weak* topology on $$X^*$$ look like? My current understanding is that they look like $$f^{-1}(U) = \Lambda_x^{-1}(U)$$ for any open set $$U \subset \mathbb{F}$$. Is this correct?

@cbbam this is too long for a comment the elements of the dual $$X^{*}$$ are already "there",i.e $$X^{*}$$ consists of all continuous linear forms $$f:X\to \mathbb{K}$$,this is already a normed space whose elements are forms,picking (a fixed) $$x$$, the map (evaluation at $$x$$) $$ev_{x}:X^{*}\to \mathbb{K};f\to f(x)$$ is continous,i.e $$ev_{x}\in X^{**}$$ (all this is proved befor introducing any weak or weak* topology),in fact the map $$i:X\to X^{**},x\to ev_{x}$$ is the natural embedding of $$X$$ into $$X^{**}$$ and a space is called reflexive if this embedding is surjective,So u can see $$x$$ as a vector on which elements of $$X^{*}$$ acts or as linear form acting on elements of $$X^{*}$$.Let's return now to the weak* topology on $$X^{*}$$,this is by definition the smallest (coresest) topology on $$X^{*}$$ making all $$ev_{x}$$ continous,So we want all our $$ev_{x}$$ to be continous but with the least amount of open subsets,this means that given an open subset $$U\subset \mathbb{K}$$,$$ev_{x}^{-1}(U)$$ must be open for every $$x\in X$$,Now $$ev_{x}^{U}=\{f\in X^{*} \backslash f(x)\in U\}$$, so an open subset of $$X^{*}$$ consists of all (continous since this is the definition of $$X^{*}$$) linear form that send ( a fixed) $$x$$ inside $$U$$.In general if you have topological space $$X, Y$$ and u pick a family of $$f_{i}\in C(X,Y)$$,then u can define a new topology on $$X$$ a the smallest topology on $$x$$ making all the $$f_{i}$$ continuous,how does open subsets of this new toplogy looks like?well let $$U_{k},k=1,..n$$ be open subsets of $$Y$$,a basic open subset has the form $$\cap_{k=1}^{n}f_{i_{k}}^{-1}(U_{k})$$,when u take $$Y=\mathbb{K}$$ and the family of continuous maps to be $$X^{**}$$, u get the weak topology on $$X^{*}$$,which as I mentioned before coarser that the weak* topology(remark here we require all linear forms on $$X^{*}$$ not only $$ev_{x}$$,that why we get "more" open subsets in general). u can look for Initial topology if you are intersted, hope this helps.
• Thank you this is very helpful! I have one follow up question, so we require that $\Lambda_x(f) = f(x)$ be continuous. Does this require that $f$ also be continuous for all $y \neq x$? Commented Nov 4, 2022 at 17:42
• @CBBAM yes $f$ is an element of $X^{*}$, so (by definition) it is continous at every point of $X$.remember that we trying to put a topology on $X^{*}$ (which already ahs its elements, we don't change their nature).in conclusion $X^{*}$ has three topologies(of course its elements are the same ,continuous linear forms $f:X\to \mathbb{K}$),one is induced by the operator norm,$\lvert \lvert f\rvert \rvert$ Commented Nov 4, 2022 at 19:33
• @CBBAM The second one is the topology making every evaluation continuous,where an open subset is given by $ev_{x}^{-1}(U)$,that is all linear forms $f$ that send a fixed vector $x$ inside an open subset $U$ (and is comming from $X$)and the one making every linear form on $X^{*}$,that is every element of $X^{**}$ continous ,the first one is the finer (with the most amount of open subsets ),while the second one is the coarser(with the least amount of open subsets) Commented Nov 4, 2022 at 19:34
it is an open subset of $$X^{*}$$, so it need to consisit of linear forms, more precisely it consists of $$\lambda_{x} ^{-1}(U)=\{f\in X^{*} \backslash f(x) \in U \}$$ ($$x$$ is fixed and the open subset consists of forms that send $$x$$ inside $$U$$),reamark that if u restrict to the evaluations $$\lambda_{x}:X^{*}\to \mathbb{F},\lambda_{x}(f)=f(x)$$ then u get the weak* topology on $$X^{*}$$,which is coarser than the weak topology comming from the double dual,unless $$X$$ is reflexive.
• I see, so here we require $f$ to be continuous for only some $x$ (namely $f(x)$), which in turn makes $\Lambda_x$ continuous for every $f$? Commented Nov 4, 2022 at 16:48