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Let us consider the (metric) theory of Banach algebras. I have a sentence encoding the (possible) openness of multiplication in a given Banach algebra:

$$(\forall x) (\forall y) (\forall \varepsilon > 0)( \exists \delta > 0)(\forall z)\; \big(\|z-xy\| < \delta \Rightarrow (\exists u)(\exists v)[\|u-x\|<\varepsilon \& \|v-y\|<\varepsilon \;\&\; z = uv]\big)$$

This sentence looks quite first-order to me.

Suppose that a Banach algebra $A$ satisfies the above formula. Can we directly conclude that some ultrapower of $A$ would also satisfy it?

By an ultrapower, I mean the metric (Banach-space) ultrapower. I know that people in C*-algebras have mastered such methods but I am not sure to what extent a similar machinery is available in the more general setting of Banach algebras.

(I know that this holds for all ultrapowers if we shift $(\forall x)(\forall y)$ after $(\exists \delta > 0)$ but the reason is not model-theoretic and the condition is too strong for me.)

Should it hold, I would be most grateful for pointing out the relevant literature touching this topic.

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  • $\begingroup$ Generally speaking an open condition like $\exists \delta > 0$ is not going to be preserved under taking ultrapowers unless there is some positive $\gamma > 0$ such that $\delta$ is actually always $\geq \gamma$. $\endgroup$ – James Hanson Jul 28 '18 at 15:33
  • $\begingroup$ Although in this case since you're also quantifying over $\varepsilon$, it's a little more complicated, but my comment was getting too long so I'll write it as an answer. $\endgroup$ – James Hanson Jul 28 '18 at 15:35
  • $\begingroup$ It is a first order sentence, but the problem is that some variables are elements of the Banach algebra and others are real numbers. So it's first order in a two-sorted language. That means that general nonsense only tells you the sentence will still be true in an ultrapower with $\epsilon$ and $\delta$ referring to elements of an ultrapower of $\mathbb{R}$. $\endgroup$ – Nik Weaver Jul 28 '18 at 15:52
  • $\begingroup$ (So I assume you cannot conclude this for ultrapowers, but I don't have a counterexample.) $\endgroup$ – Nik Weaver Jul 28 '18 at 15:53
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    $\begingroup$ @JamesHanson: Yes, I know. Okay, so in the unmodified ultrapower you'd expect the formula to hold with infinitesimal $\delta$, but after factoring out null vectors you'd expect it to simply fail. $\endgroup$ – Nik Weaver Jul 28 '18 at 16:27
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What makes this tricky is that metric or continuous logic (in the sense of this document) doesn't directly allow for quantification over $\varepsilon$'s and $\delta$'s like that. You can still express it with a schema, but what's difficult in this case is that you have quantification over elements of the structure before quantifying over $\varepsilon$ and $\delta$. There's also some subtlety with how you formalize implication in this context.

Generally speaking a statement is only going to be preserved by ultrapowers/metric elementary equivalence if it's uniformly true in the structure. This fact is baked into the formalism in some places, such as the requirement to give a modulus of uniform continuity for a given function. If a function in a metric structure fails to be uniformly continuous, then in some elementary extension it won't even be a function. More generally purely 'topological' properties are often too fragile to be preserved under ultrapowers unless they are actually true in some uniform way.

EDIT2: This is not as straightforward as I thought. I can tell you roughly when this is not going to work. If there is an $\varepsilon > 0$ such that for every $\gamma > 0 $ there exists vectors $x,y$ with $\| x\| = \| y\|=1$ such that the $\delta$ needed to satisfy your condition is $<\gamma$, then in an ultrapower (over a countable index set) the condition will fail. So I would guess that in order for the condition to be satisfied in ultrapowers of the structure you actually need the stronger form you mentioned where the $(\forall x)(\forall y)$ quantifiers are after the $(\exists \delta > 0)$ quantifier. And I mean 'need' in a strong sense, i.e. if you have a Banach algebra such that your condition is true in all ultrapowers then the stronger form of the condition (although restricted to norm 1 vectors) actually holds in the first place.

EDIT: I think that maybe you can construct a counterexample from the proof of theorem 4.12 here, specifically some kind of product algebra of the algebras $\ell_1(\mathbb{Z}_{n!})$ with the convolution algebras. EDIT3: This reference is still relevant but I was being too optimistic about openness of multiplication being preserved under products.

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  • $\begingroup$ "If there is an $\epsilon > 0$ ... the condition will fail." Are you sure? What would be the falsifying $x$, $y$, and sequence of $z$'s? I think this is more subtle than you think ... $\endgroup$ – Nik Weaver Jul 28 '18 at 16:31
  • $\begingroup$ There's a lot of quantifiers, so it's entirely possible that it's more subtle than I think. If I'm not mistaken the original condition is equivalent to the following $(\forall x)(\forall y)(\forall\varepsilon>0)(\exists\delta>0)(\forall z)(\exists u)(\exists v)(\|z-xy\|\geq\delta\vee[\|u-x\|\leq\varepsilon\wedge\|v-y\|\leq\varepsilon\wedge z=uv])$ where now I'm restricting to quantification over norm 1 vectors (other than the quantification for z, u, and v). $\endgroup$ – James Hanson Jul 28 '18 at 16:53
  • $\begingroup$ What I want to say is that if I let $\varphi(x,y,\varepsilon,\delta) = (\forall z)(\exists u)(\exists v)(\|z-xy\|\geq\delta\vee[\|u-x\|\leq\varepsilon\wedge\|v-y\|\leq\varepsilon\wedge z=uv])$, then for fixed inputs this is preserved under passing to a modified ultrapower. $\endgroup$ – James Hanson Jul 28 '18 at 16:53
  • $\begingroup$ So then if I let $D(x,y,\varepsilon)=\{\delta:\varphi(x,y,\varepsilon,\delta)\}$, then if there exists a sequence $\{x_n,y_n\}$ and an $\varepsilon>0$ such that $\sup D(x_n,y_n,\varepsilon )\rightarrow 0$ as $n\rightarrow \infty$, then in any nontrivial ultrapower of the Banach algebra over a countable index set, the limits of $x_n$ and $y_n$ will be counterexamples. $\endgroup$ – James Hanson Jul 28 '18 at 16:54
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    $\begingroup$ James, this will be published (not by me) but BV is indeed a counterexample. $\endgroup$ – Tomek Kania Jul 28 '18 at 19:35

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