In my paper [On the optimal error bound for the first step in the method of cyclic alternating projections](https://arxiv.org/abs/1908.00531), I defined functions $f\_n:[0,1]\to\mathbb{R}$,
$n\geqslant 2$, by
$$
f\_n(c)=\sup\{\|P\_n\dotsm P\_2 P\_1-P\_0\|\,|\,c\_F(H\_1,\dotsc,H\_n)\leqslant c\},
$$
where


(1) the supremum is taken over all complex Hilbert spaces $H$ and systems of closed subspaces $H\_1,\dotsc,H\_n$ of $H$ such that the Friedrichs number $c\_F(H\_1,\dotsc,H\_n)$ is less than or equal to $c$. The Friedrichs number is a quantitative characteristics of a system of closed subspaces,
$c$ is a given number from $[0,1]$.


(2) $P\_i$ is the orthogonal projection onto $H\_i$, $i=1,2,\dotsc,n$ and $P\_0$ is the orthogonal projection onto the subspace $H\_1\cap H\_2\cap\dotsb\cap H\_n$.


**Now I have some doubts in the validity of this definition.**
Namely, I know that all sets do not constitute a set and if I understand things right, all Hilbert spaces do not constitute a set (and even all one dimensional Hilbert spaces do not constitute a set). **So, can I take the supremum?**


**Please help me; I will be very grateful for any comments, remarks, and answers.**


In response to Mike Miller's [comment](https://mathoverflow.net/questions/375759/can-we-take-a-supremum-over-all-hilbert-spaces#comment952651_375759):
Unfortunately, I also know almost nothing about set theory.
Yes, I understand that in fact I take the supremum of a "set" of real numbers, namely, of the "set" $A=A\_n(c)$
which consists 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
$\|P\_n\dotsm P\_2 P\_1-P\_0\|=a$.
**But I do not understand why the "set" $A$ is a set.**



Can we take a supremum over all Hilbert spaces?
It is true that we cannot use an arbitrary property$^1$ $P$ to define a set, in the sense that the collection of **all** things with property $P$ need not be a set. However, the [axiom (scheme) of separation](https://en.wikipedia.org/wiki/Axiom_schema_of_specification) says that we **can** use an arbitrary property to define a *subset*: whenever $X$ is a set, the collection $\{x\in X: P(x)\}$ is also a set.


So just take $X=\mathbb{R}$ and $P(x)$ = "There is a Hilbert space such that …". Per Separation, we get that your collection of reals $A$ is in fact a set. And we may now take its supremum.


Note that this illustrates an important point about how $\mathsf{ZFC}$ (and its variants) get around Russell's paradox:



> 
> It's **size**,$^{2}$ not **complexity of definition**, which controls whether or not a collection is a set or a proper class in $\mathsf{ZFC}$.$^{3}$
> 
> 
> 


Part of the success of $\mathsf{ZFC}$ is due to the ease with which we can in fact verify that something is a set. The only time you'll run into trouble is when you want to form a set which isn't a priori part of some bigger thing you already know is a set; here we may have to think a bit (although the [axiom (scheme) of *replacement*](https://en.wikipedia.org/wiki/Axiom_schema_of_replacement) similarly makes things usually very easy, once it's mastered).




---


EDIT: Per the [comments below](https://mathoverflow.net/questions/375759/can-we-take-a-supremum-over-all-hilbert-spaces#comment953257_375762), let me sketch how to define "complete metric space" in the language of set theory. As you'll see, even the sketch is quite lengthy; if there's a particular point you'd like further information on, I suggest asking a separate question at MSE.


Here's the sequence of definitions we need to whip up:


1. We need to talk about ordered pairs, functions, and Cartesian products.
2. We need to build $\mathbb{N}$, so that we can build $\mathbb{Q}\_{\ge 0}$, so that we can build $\mathbb{R}\_{\ge 0}$; along the way we'll need the notions of equivalence relation and equivalence class, of course.
3. While the previous two points will be enough to define metric spaces ("An ordered pair $(X,\delta)$ where $X$ is a set and $\delta:X^2\rightarrow\mathbb{R}$ such that [stuff]"), to define *complete* metric spaces we'll also need the notions of **infinite sequence** and **equivalence relation/class**.


The first bulletpoint is standard set-theoretic fare which you'll see treated in the beginning of any text on set theory, so I'll skip it; if you're interested, though, you can start with [the wiki page on ordered pairs](https://en.wikipedia.org/wiki/Ordered_pair).


The third is really the first in disguise: an infinite sequence is just a function with domain $\mathbb{N}$.


So all the "meat" is in bulletpoint 2. We proceed as follows:


* First, we'll use the von Neumann approach to $\mathbb{N}$: an *ordinal* is a hereditarily transitive set, ordinals are ordered by $\in$, and the finite ordinals are the ordinals which do not contain any (nonempty) limit ordinal. We then identify $\mathbb{N}$ with the finite ordinals — more jargonily, $\mathbb{N}=\omega$. We define addition and multiplication of ordinals via transfinite recursion as usual.
* Next, we consider the equivalence relation $\sim$ on $\omega\times(\omega\setminus\{0\})$ as follows: $$\langle a,b\rangle\sim\langle c,d\rangle \iff ad=bc,$$ and we let $\mathbb{Q}\_{\ge0}$ be the set of $\sim$-classes. We lift the ordering on $\omega$ to $\mathbb{Q}\_{\ge 0}$ in the obvious way.
* Now we're ready to define $\mathbb{R}\_{\ge 0}$, via Dedekind cuts: an element of $\mathbb{R}\_{\ge 0}$ is a nonempty, downwards-closed, bounded-above subset of $\mathbb{Q}\_{\ge0}$. The ordering on $\mathbb{R}\_{\ge 0}$ is just $\subseteq$.


With all this in hand, the naive definitions of metric space, Cauchy sequence, and complete metric space translate into the language of set theory directly (if tediously). The point is that all of this is first-order *in set theory*, with axioms like Powerset (which, despite what they mean intuitively, are indeed first-order) doing the heavy lifting needed to show that the objects we want actually exist at all. (For a bit more about the nuance of "first-order in set theory," see [this recent answer of mine](https://math.stackexchange.com/a/3898216/28111).)




---


$^1$Really I mean "first-order formula," but I don't want to get too much into the details.


$^{2}$Specifically, in a precise sense we have: a class is a proper class iff it surjects onto the class of ordinals. This is not the same as the [principle of limitation of size](https://en.wikipedia.org/wiki/Axiom_of_limitation_of_size), but it's of similar flavor.


$^3$I should observe that this isn't the only possible response to the need to distinguish between sets and proper classes: there are other set theories (e.g. $\mathsf{NF}$, $\mathsf{GPK^+\_\infty}$, ...) which take the other approach. However, these theories make it harder to check whether something is in fact a set.


