Can we take a supremum over all Hilbert spaces?

Question

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.

Answer 1

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 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 similarly makes things usually very easy, once it's mastered).

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EDIT: Per the comments below, 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.

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.)

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$^1$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, 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.

Answer 2

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.