0
$\begingroup$

Let $n$ be a fixed natural number. 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).

As explained in a previous question, there is no problem with the definition of the supremum.

The question is about the following argument, which seemingly requires the axiom of choice for proper classes.

By the definition of the supremum for every natural number $m\geqslant 1$ there exist a Hilbert space $H^{(m)}$ and a system of its closed subspaces $H^{(m)}_1,...,H^{(m)}_n$ such that $c_F(H^{(m)}_1,...,H^{(m)}_n)\leqslant c$ and $\|P^{(m)}_n...P^{(m)}_2 P^{(m)}_1-P^{(m)}_0\|>f_n(c)-1/m$. Now we can use these systems of subspaces, for example, we can form the orthogonal direct sum $\bigoplus_{m=1}^\infty H^{(m)}$.

Question: are these innocent arguments correct, say, in the axiomatic theory ZFC? I am suspicious here because I need to choose infinite number of systems of subspaces simultaneously. If I understand things correctly, we need to use here the Axiom of Choice. But all systems of subspaces $(H;H_1,...,H_n)$ with $c_F(H_1,...,H_n)\leqslant c$ do not form a set and the Axiom of Choice works with sets! So we need to use here something like the "Axiom of Choice for classes", but I do not know such an Axiom. Therefore I think that the arguments above are not correct, but they are so innocent...

Help me please understand if the arguments are correct or not correct. In other words, can we choose a sequence of systems of subspaces of Hilbert spaces as above or we cannot do this?

$\endgroup$
13
  • 11
    $\begingroup$ You've asked this several times already. You've received the same answer several times already. It will be very helpful if you'll clarify what exactly is missing at this point. You're not actually choosing uniformly. Think of each $H$ as a room: you first enter the room, then you have a lot of rooms for the choice of subspaces, so you enter into one of the rooms. You compute something and you get a real number. You keep that real number, and you now use Replacement to instantly do the same for all rooms and all the rooms inside of those rooms. You are not choosing anything. $\endgroup$ – Asaf Karagila Apr 19 at 18:23
  • 4
    $\begingroup$ I'm sorry if I don't see it Ivan, but to me it looks almost the same proof that you're asking about mathoverflow.net/questions/375759 and mathoverflow.net/questions/387353 and mathoverflow.net/questions/389610, where is the difference? $\endgroup$ – Asaf Karagila Apr 19 at 18:44
  • 5
    $\begingroup$ I think I understand the issue. You seem to think that $\forall x\exists y\varphi(x,y)$ is logically equivalent to the $\exists F\forall x\varphi(x,F(x))$ (here $F$ is a 2nd order object), and while that is true that the two are equivalent in $\sf ZF+GC$, this is not a rule of logic, nor we use this in the proof. So there's no need to choose for all Hilbert spaces such system of closed subspaces. So we are not doing that. $\endgroup$ – Asaf Karagila Apr 19 at 18:55
  • 4
    $\begingroup$ You can show the existence of a choice function for a countable (or more generally, set-indexed) family of nonempty classes by employing Scott’s trick which reduces it to a family of sets, and then you can apply the usual axiom of choice. $\endgroup$ – Emil Jeřábek Apr 19 at 19:30
  • 2
    $\begingroup$ @AsafKaragila: The answer to this question is quite different from the answers to previous questions. The previous questions did not require Scott's trick, and this question does. $\endgroup$ – Dmitri Pavlov Apr 19 at 22:09
3
$\begingroup$

For every $m\ge1$ we have a proper class $A_m$ of tuples $(H,H_1,…,H_n)$ satisfying the described condition.

Now invoke Scott's trick and construct a subset $A'_m⊂A_m$ for each $m\ge1$, consisting of tuples with the minimal rank in $A_m$. (This part of the construction does not use the axiom of choice.)

Since $A'_m$ are ordinary sets, we can apply the ordinary axiom of choice to extract a specific tuple for each $m\ge1$.

As a side remark, Scott's trick is also used in various constructions with classes, such as quotients of classes, and, more generally, colimits of classes. See the nLab article category of classes for more information.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.