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In our functional analysis lecture we defined the weak topology in a what seems to me like a non canonical way, i.e. not as unions of finite intersections of preimages of open sets in the underlying field (we only use real or complex numbers) under continuous functionals. At the moment I work with the definition above and have trouble bringing it together with the definitions we used in the lecture. Mainly because those are a mix of informal speak and quantifiers, where I'm not even sure whether they were used right.

Our definition goes as follows: A set A is in the weak topology, iff $$\forall a \in A \exists\varepsilon\exists \phi_1,...,\phi_n \in V^*: \{y: \lvert\phi_i(a-y)\rvert<\varepsilon\}\subseteq A$$

I can't work with this 'informally formal' definition. Apart from n being used without being defined first (it obviously has to be a natural number), I can't make sense of the finite list of continuous functionals. If all we wanted to do is to make sure there exists any such functional at all, there wouldn't be any reason to demand a list. On the other hand, if the expression stays like this, there still is the possibility that more than the listed functionals with the same property as those exist. It's only an existential quantifier after all. If I take the next logical step, I would have to conclude that the (formally) right definition would be

A set A is in the weak topology, iff $$\forall a \in A\exists\varepsilon\in\mathbb{R}_{>0} \exists n \in \mathbb{N}\exists^n \phi \in V^*: \{y: \lvert\phi(a-y)\rvert<\varepsilon\}\subseteq A$$

Another definition we used in the lecture is: Open sets in the weak topology are unions and finite intersections of sets of the form $\{x: a<\lvert \phi(x)\rvert<b\}$ for a,b>0 and $\phi\in X^*$. Again, what $\phi$? All? Any? Exactly one? What is with the a and b, do they depend on the x? I can't work with this.

Help would really be appreciated. Thank you in advance.

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    $\begingroup$ Where you wrote $M$ in your definitions should that be $A$? $\endgroup$ – Rudy the Reindeer Aug 29 '15 at 0:20
  • $\begingroup$ Yes. Thank you for pointing it out. $\endgroup$ – azureai Aug 29 '15 at 18:19
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The precise statement is the following: a set $A\subseteq V$ is in the weak topology iff $$\forall a\in A \exists \epsilon\in \mathbb{R}_{>0}\exists F\subseteq V^*: F\text{ is finite and }\{y\in V\mid\forall\phi\in F:|\phi(a-y)|<\epsilon\}\subseteq A.$$

It sounds like the key point you are missing is the implicit universal quantifier in the definition of the set which is supposed to be contained in $A$: the ambiguous expression $\{y:|\phi_i(a-y)|<\epsilon\}$ is supposed to refer to the set of $y$ such that $|\phi_i(a-y)|<\epsilon$ for all $i$ (i.e., for $i=1,\dots n$). In relation to the definition you do understand, this "for all" corresponds to taking a certain finite intersection of preimages of open sets under continuous functionals, namely the intersection of the finitely many sets $\{y:|\phi_1(a-y)|<\epsilon\}, \{y:|\phi_2(a-y)|<\epsilon\},\dots, \{y:|\phi_n(a-y)|<\epsilon\}$

As for the second definition, a set $A$ is "of the form $\{x:a<|\phi(x)|<b\}$" if there exist $a,b>0$ and $\phi\in X^*$ such that $A=\{x:a<|\phi(x)|<b\}$. A set $B$ is "a finite intersection of sets of the form $\{x:a<|\phi(x)|<b\}$" if there exists a finite collection $F$ of sets such that $V=\bigcap_{A\in F} A$ and every element of $F$ is of the form $\{x:a<|\phi(x)|<b\}$. Note that this means that the $a$, $b$, and $\phi$ in question are different for each element of $F$; you should think of the entire prhase "of the form $\{x:a<|\phi(x)|<b\}$" as simply being an adjective. Finally, a set $C$ is open in the weak topology if there exists a collection $\mathcal{B}$ of sets such that $C=\bigcup_{B\in\mathcal{B}} B$ and every element of $\mathcal{B}$ is a finite intersection of sets of the form $\{x:a<|\phi(x)|<b\}$. Here again, you should think of "a finite intersection of sets of the form $\{x:a<|\phi(x)|<b\}$" as being an adjective, so the finite collection $F$ involved is different for each $B\in\mathcal{B}$.

More generally, to say that an object $A$ is "of the form blah" where "blah" includes some unspecified variables means that there exist values you can plug in for the variables such that "blah" then evaluates to $A$.

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  • $\begingroup$ Thank you for the elaboration, the definition makes sense to me. However, I still can't quite understand how the first definition in my opening post implies the corrected version which was first in your post. The other way is trivial. $\endgroup$ – azureai Aug 30 '15 at 19:17
  • $\begingroup$ The idea is sketched in TrialAndError's answer. In more detail, a basis for the first definition is given by sets of the form $U=\bigcap_{i=1}^n \phi_i^{-1}(U_i)$, for functionals $\phi_i$ and open sets $U_i$. Given any $a\in U$, we can find an $\epsilon>0$ such that for every $i$, $B_\epsilon(\phi_i(a))\subseteq U_i$. But $\phi_i$ is linear, so $\phi_i^{-1}(B_\epsilon(\phi_i(a)))=\{y:| \phi_i(a)-\phi_i(y)|<\epsilon\}=\{y:|\phi_i(a-y)|< \epsilon\}$. $\endgroup$ – Eric Wofsey Aug 30 '15 at 19:35
  • $\begingroup$ Neat. And to obtain the definition with the sets of form $C=\{x:a<\lvert \phi(x)\rvert <b\}$ we just assume w.l.o.g. $\varepsilon <\lvert \phi(x) \rvert$, right? $\endgroup$ – azureai Aug 30 '15 at 22:19
  • $\begingroup$ Er, for that version you actually need to drop the absolute value (and allow $a$ and $b$ to be negative), so you're talking about $\{x:a<\phi(x)<b\}$ (if you work with complex scalars, you should also replace $\phi$ with $\operatorname{Re} \phi$). If you don't make any such modification, for instance, no such set can contain the vector $0$, and every open set containing $x$ would also have to contain $-x$. With such a modification, it is easy to see that it gives the same topology. $\endgroup$ – Eric Wofsey Aug 30 '15 at 22:58
  • $\begingroup$ Right, I mixed the inclusion up. What do you mean by dropping the absolute value and replacing $\phi$ by the real part, it's both part of the definition isn't it. And one other thing: The definition as I have it on paper says 'unions and finite intersections' not 'unions of finite intersections', is this an error in the lecture? $\endgroup$ – azureai Aug 30 '15 at 23:44
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The weak topology $\tau_{w}$ is the weakest topology for which the elements of $V^{\star}$ are continuous from $(X,\tau_{w})$ to the scalar field.

For any $\phi \in V^{\star}$, the sets $\phi^{-1}B_{r}(0)$ must be open neighborhoods of $0$ for any $r > 0$. Finite intersections of such inverse images under elements of $V^{\star}$ will form a base of neighborhoods at $0$. Neighborhoods of other non-zero points of $X$ are translations of the neighborhood base at $0$ because of linearity.

Using these basic facts gives you the definition of a general open set.

The interesting question is this: What are the continuous linear functionals from $(X,\tau_{w})$ into the scalar field? It's a great question with a nice answer.

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  • $\begingroup$ -1 vote: I guess some are not familiar with how to create the weakest topology for which a set of functions is continuous. en.wikipedia.org/wiki/Weak_topology $\endgroup$ – DisintegratingByParts Aug 29 '15 at 9:39
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    $\begingroup$ This answer doesn't actually address the question that was asked, is unlikely to be very helpful to the asker, and honestly makes it look like you didn't even read the body of the question. $\endgroup$ – Eric Wofsey Aug 29 '15 at 14:43
  • $\begingroup$ @EricWofsey : That's your opinion. But I do in fact address the issues by looking at a higher level. First, I explain how the topology arises, and the necessity of the definition from the point of view of creating the weakest topology for which a collection of linear functionals is continuous with respect to the given topology. Second, it should then be clear why one must use finite intersections of sets $\phi^{-1}B_{r}(0)$. Finally, translation invariance becomes a non-issue; because of the linearity of functionals, the neighborhood base may be translated to other non-zero points by addition. $\endgroup$ – DisintegratingByParts Aug 29 '15 at 16:34
  • $\begingroup$ @EricWofsey : My goal was to help the OP unravel the definitions by seeing how they arise, and why they are the way they are. This reduces the number of quantifiers as well. Sometimes it's better to step back a bit and not look at things so technically. $\endgroup$ – DisintegratingByParts Aug 29 '15 at 16:36
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    $\begingroup$ Nice, got it. Thank you. $\endgroup$ – azureai Aug 30 '15 at 22:23

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