TITLE: Can we take a supremum over all Hilbert spaces?
QUESTION [19 upvotes]: In my paper On the optimal error bound for the first step in the method of cyclic alternating projections, 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:
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.

REPLY [6 votes]: The purpose of this answer is to recall that the classical proof of the existence of suprema of bounded families of reals (as found in any analysis textbook) continues to work without any modifications whatsoever for proper classes.  (In particular, we do not assume that the existence of suprema of bounded nonempty subsets of reals has been proved yet.)
In fact, the proof works in the Zermelo set theory,
without the axiom of replacement and the axiom of choice.
Theorem.
Suppose $C$ is a nonempty class (possibly proper) and $g\colon C\to \mathbf R$ is a map bounded from above.
Then $\sup_{c\in C} g(c)$ exists.
Proof.
Denote by $U$ the set of upper bounds for $g$,
i.e., $U=\{u\in\mathbf R\mid \forall c\in C\colon g(c)\le u\}$,
which exists by the axiom of separation.
Since $\sup_{c\in C} g(c)$ is by definition the smallest upper bound for $g$,
we have to show that $U$ has the smallest element.
By assumption, $U$ is nonempty.
Since $C$ is nonempty, the set $U$ is bounded from below,
namely, it is bounded from below by $g(c)$ for any $c\in C$.
Thus, $U$ is a nonempty upward-closed subset of reals bounded from below,
i.e., an upper Dedekind cut, so $\inf U$ exists.
It remains to show that $\inf U \in U$.
Indeed, $\inf U \in U$ is equivalent to
$$\forall c\in C\colon g(c) \le\inf U,$$
which in its turn (by definition of $\inf$) is equivalent to
$$\forall c\in C \forall u\in U\colon g(c) \le u,$$
equivalently,
$$\forall u\in U \forall c\in C\colon g(c) \le u,$$
i.e.,
$$\forall u\in U\colon u\in U,$$
which is tautologically true.
Thus, $\inf U \in U$, so $\inf U$ is the smallest upper bound for $g$, which, therefore, exists.