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This question is related to my question Can we choose an element from a class?. However, I decided to create a separate question.

Let $H$ be a complex Hilbert space and $H_1,\dotsc,H_n$ be closed subspaces of $H$. Set $H_0\mathrel{:=}H_1\cap H_2\cap\dotsb\cap H_n$ and let $P_i$ be the orthogonal projection onto $H_i$, $i=0,1,2,\dotsc,n$. I study the functions $f_n:[0,1]\to\mathbb{R}$ defined by $$ f_n(c)=\sup\{\lVert P_n\dotsm P_2 P_1-P_0\rVert \mathrel| c_F(H_1,\dotsc,H_n)\leqslant c\},\,c\in[0,1], $$ where the supremum is taken over all complex Hilbert spaces $H$ and systems of closed subspaces $H_1,\dotsc,H_n$ of $H$ for which the Friedrichs number $c_F(H_1,\dotsc,H_n)$ is less than or equal to $c$ (the Friedrichs number is a certain numerical characteristic of a system of subspaces). Note that all such systems of subspaces do not form a set. Despite this, the function $f_n$ is well-defined (see my Question Can we take a supremum over all Hilbert spaces?). Indeed, let $A_{n}(c)$ be the set of all $a\in\mathbb{R}$ for which there exist a complex Hilbert space $H$ and a system of closed subspaces $H_1,\dotsc,H_n$ of $H$ such that $c_F(H_1,\dotsc,H_n)\leqslant c$ and $\lVert P_n\dotsm P_2 P_1-P_0\rVert=a$. Then by the axiom (scheme) of separation $A_{n}(c)$ is a set and thus we can take its supremum.

I need to show that $f_n(c)\leqslant g_n(c)$ for some function $g_n$. I argue as follows. Consider arbitrary element $a\in A_{n}(c)$. Then there exist a complex Hilbert space $H$ and a system of closed subspaces $H_1,\dotsc,H_n$ of $H$ such that $c_F(H_1,\dotsc,H_n)\leqslant c$ and $\lVert P_n\dotsm P_2 P_1-P_0\rVert=a$. After this I work with this system of subspaces $(H;H_1,\dotsc,H_n)$ and show that $\|P_n\dotsm P_2 P_1-P_0\|\leqslant g_n(c)$. Thus $a\leqslant g_n(c)$. Since this inequality holds for every $a\in A_{n}(c)$, we conclude that $\sup A_{n}(c)\leqslant g_n(c)$, i.e., $f_n(c)\leqslant g_n(c)$.

Questions. Are all these arguments correct, say, in the axiomatic theory ZFC? Essentially, the core of my worries is the following. Unfortunately, I do not understand if the function $A_n(c)\ni a\mapsto (H;H_1,...,H_n)$ such that $c_F(H_1,\dotsc,H_n)\leqslant c$ and $\lVert P_n\dotsm P_2 P_1-P_0\rVert=a$, is needed in the arguments above or not. If a function is needed here, it turns out that the "Axiom of Choice for classes" is needed here, and I do not know what to do with this. On the one hand, it seems that the function is not needed here. On the other hand, we choose a system of subspaces for each $a\in A_n(c)$; the set $A_n(c)$ can be infinite and we need to consider all $a\in A_n(c)$. Therefore, perhaps, a function is needed here. Explain to me, please, whether a function is needed here or not?

Please help me.

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    $\begingroup$ As in all your questions, you do not need the axiom of choice to choose one element from a non-empty class; that's what non-empty means. You only need AC to make many choices simultaneously. See, for example, mathoverflow.net/questions/387353/… . $\endgroup$
    – LSpice
    Apr 7 at 21:47
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    $\begingroup$ Also, a TeX note: please do not use repeated periods to simulate dots; that's what the various \dots commands are for. Compare the spacing in $H_1 \cap H_2 \cap \dotsb \cap H_n$ H_1 \cap H_2 \cap \dotsb \cap H_n to that in $H_1 \cap H_2 \cap ... \cap H_n$ H_1 \cap H_2 \cap ... \cap H_n. I have edited accordingly. $\endgroup$
    – LSpice
    Apr 7 at 21:53
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    $\begingroup$ The answers are as follows: these arguments are correct, the indicated function is not needed. Indeed, your argument might as well prove that for all H1,…,Hn, H such that c_F(H1,…,Hn)⩽c, you have ∥Pn⋯P2P1−P0∥⩽g_n(c). Quantifying over a collection of sets is perfectly legitimate even if this collection of sets forms a proper class. There is no need to mention A_n(c) at all, it only creates further confusion. $\endgroup$ Apr 7 at 22:57
  • $\begingroup$ @DmitriPavlov Would you mind posting your comment as an answer to close out the question? $\endgroup$
    – Alec Rhea
    Apr 8 at 0:25
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The answers are as follows: these arguments are correct, the indicated function is not needed.

Indeed, your argument might as well prove that for all $H_1,\dotsc,H_n$, and $H$ such that $c_F(H_1,\dotsc,H_n)\le c$, you have $\lVert P_n\dotsm P_2P_1−P_0\rVert \le g_n(c)$. Quantifying over a collection of sets is perfectly legitimate even if this collection of sets forms a proper class. There is no need to mention the set $A_n(c)$ at all, it only creates further confusion.

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  • $\begingroup$ Dear Dmitri, thank you for your answer. One thing is confusing for me. Assume that my arguments (proof of $f_n(c)\leqslant g_n(c)$) are correct. Then by "similar" arguments I can prove the Axiom of Choice. The "proof" is as follows. Let $A_i, i\in I$ be a set of mutually disjoint nonempty sets. We will construct a set that contains exactly one element in common with each of the sets as follows. Consider arbitrary $i\in I$ and fix it. Since the set $A_i$ is nonempty, $\exists x$ such that $x\in A_i$. To be continued. $\endgroup$ Apr 9 at 21:29
  • $\begingroup$ Now we use the existential instantiation and write $x(i)$ for a new symbol (just a symbol) such that $x(i)\in A_i$. Consider one-element sets $\{x(i)\}$ and let $A=\bigcup_{i\in I}\{x(i)\}$. Then $A$ is a set and $A$ contains exactly one element in common with each $A_j$. Where am I wrong? $\endgroup$ Apr 9 at 21:37
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    $\begingroup$ @IvanFeshchenko You go wrong as soon as you define $A$: intuitively this requires you to "unfix" your chosen element $i$. You'll see this if you sit down and try to write out your argument as a formal proof (say, in sequent calculus). Basically you have a first existential instantiation picking an $i$ and following that a second existential instantiation picking an $x$, but there isn't a logical rule letting you somehow "uniformize" this. (In fact you can think of $\mathsf{AC}$ as an additional logical rule in a precise, if technical, sense.) $\endgroup$ Apr 9 at 22:12
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    $\begingroup$ @IvanFeshchenko: Your argument is not similar. In my answer, the statement is proved for all H_i and H. In your comment, you have to choose x(i). In particular, “existential instantiation” simply does not work in “families” as described; there is no such rule in first-order logic. No such choices are made in my answer. In particular, I do not choose H_i and H for each c, but rather prove the inequality for all H_i and H. $\endgroup$ Apr 9 at 22:21
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    $\begingroup$ @IvanFeshchenko: Yes, U is a set by separation: it is constructed as {u∈R| ...}, and ... can be any first-order formula, such as the one you used. $\endgroup$ Apr 20 at 23:15

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