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The uniform boundedness principle and its corollaries from a logical point of view are statements of when one can swap quantifiers in Banach spaces. Take for instance the principle of condensation of singularities; if $V$ is a Banach space and $\{\phi_{ij} \}_{i, j \in \mathbb N} \subseteq V^*$ is a family of continuous linear functionals, then \begin{equation} (\forall i \in \mathbb N)(\exists f_i \in V)(\sup_{j \in \mathbb N} |\phi_{ij} (f_i)| = \infty) \iff (\exists f \in V)(\forall i \in \mathbb N)(\sup_{j \in \mathbb N} |\phi_{ij} (f)| = \infty). \end{equation} The forward is the non-trivial direction, and one of the standard proofs of the like use Baire category, where after one plays around with quantifiers, unions and intersections, we can show that in fact the set of $f \in V$ satisfying the property on the right is in fact generic (in a category sense) in $V$.

I was wondering if there's some kind of deeper interplay of logic, topology and functional analysis going on here, perhaps a general principle for the types of statements about Banach spaces in which we can swap quantifiers $(\forall \exists \iff \exists \forall)$. If so, perhaps this gives a logical heuristic for why the uniform boundedness principle holds; equivalence of weak boundedness and strong boundedness by itself is a surprising result. My intuition says something of the sort should exists, but I haven't been able to dig anything up on the subject.

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    $\begingroup$ Swapping quantifiers depends a lot on the statement that comes after the quantifiers, and not just on the Banach space. Even in a finite set, $\forall x \exists y \; A(x,y)$ may or may not be equivalent to $\exists y \forall x \; A(x,y)$, depending on $A$. $\endgroup$ Commented May 27, 2020 at 20:14
  • $\begingroup$ Is there a logical interpretation of other quantifier swaps in analysis, e.g. any of the more basic compactness arguments? $\endgroup$ Commented May 27, 2020 at 21:07
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    $\begingroup$ See my answer to "Techniques for reversing the order of quantifiers". While you're reading it I'll try to see if your example is also about compactness. $\endgroup$ Commented May 28, 2020 at 6:15
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    $\begingroup$ Should the last $f_i$ in the displayed formula be $f$? $\endgroup$ Commented May 28, 2020 at 6:17
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    $\begingroup$ It might be worthwhile to note that the very definition of a Banach space - i.e., completeness - can be phrased in terms of a quantifier swap: A metric space $(M,d)$ is complete if and only if, for every sequence $(x_n)$ in $M$, the following two assertions are equivalent: $\exists x \in M \; \forall \varepsilon > 0 \; \exists n_0 \; \forall n \ge n_0: \; d(x_n,x) < \varepsilon$, and $\forall \varepsilon > 0 \; \exists x \in M \; \exists n_0 \; \forall n \ge n_0: \; d(x_n,x) < \varepsilon$. (Since the second assertion is equivalent to $(x_n)$ being a Cauchy sequence.) $\endgroup$ Commented May 28, 2020 at 6:44

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