"Homomorphism" from set of sequences to cardinals?

Question

First off: I barely have any set theoretic knowledge, but I read a bit about cardinal arithmetic today and the following idea came to me, and since I found it kind of funny, I wanted to know a bit more about it.

If $A$ is the set of all real positive sequences that either converge to $0$ or diverge to $\infty$, we put an equivalence relation "$\sim$" on $A$ defined as $a \sim b$ iff $\lim \frac a b \in \mathbb R ^+$.

If $B$ is the set of all infinite cardinals, can we associate to every equivalence class $[a]$ in $A/\sim$ a cardinal $p([a])$ or to every cardinal $\lambda$ an equivalence class $[q(\lambda)]$ in such a way that the map $p: A/\sim \to B$, or $q: B \to A/\sim$ is a "homomorphism"? That is, so that we have

$$ p([a] + [b]) = p([a]) + p([b]) $$ or $$ [q(\lambda + \mu)] = [q(\lambda)] + [q(\mu)]$$

If yes, could this map even be surjective, injective or an "isomorphism"? (I don0t know how many cardinals there are of course...)

It at least superficially seems to make some sense, since for cardinals $\lambda, \mu$ we have $\lambda + \mu = \max\{\lambda, \mu\}$ and the same is true for the classes of sequences, if we order them by $a < b \Leftrightarrow \lim \frac a b = 0$

Answer 1

I think that there are a handful of points that might need clarification here.

• There is no "set of all infinite cardinals", the family of all infinite cardinals is a proper class, and if it were a set it would imply the Burali-Forti paradox
• Secondly, consider the meaning of this equivalence relation: $a\sim b$ if and only if they behave the same asymptotically, up to some finite constant. You want, now to take this and say "cardinals and sequences behave similarly" in the manner that the "stronger" wins. This intuition is not too bad, although I'd have taken multiplication and not addition. However, when you want some structure to be preserved, it might take a bit more than just "behaving the same", because certain interpretation of the same idea could have very different properties, as in my next point
• Cardinals are well ordered. There is a minimal cardinal, and there is a next one after that, and a next one after that. In your equivalence classes take $n^{-x}$ and $n^{-y}$, for choice of different $x,y\in\mathbb{R}^+$ those will give you different equivalence classes, it means you have continuum many equivalence classes and in terms of ordering - you're a lot like the order of $\mathbb{R}^{\ge 0}$ which is dense, which makes the idea of taking a cardinal for each equivalence class (even if you do only limit yourself to the first continuum-many cardinals) a tad problematic.
• And another point would be infinite summations. Cardinal arithmetic is slightly different when it comes to infinite sums, for example take the first $\aleph$ number which has an uncountable index (denoted by $\aleph_{\omega_1}$) then every addition of the form $\sum_{i\in\mathbb{N}} \lambda_i$ where all the $\lambda_i<\aleph_{\omega_1}$ is still a lot smaller than $\aleph_{\omega_1}$. So unlike the real numbers, adding uncountably many cardinals would have meaning if the Continuum Hypothesis fails,
• If however CH holds, and you only speak of the first $\omega_1$ cardinals then summation has a different trick up its sleeve: $\sum_{i\in\mathbb{N}}\lambda_i$ is just the biggest $\lambda_i$ or the first one which is bigger than all of them. I'm not quite certain how that would be handled by the equivalent classes.

That been said, "homomorphism" is not dealing with infinite summations so the two points about that might be redundant, but these are things to consider when you go between two structures, especially infinite ones that have a very strong meaning when it comes to sequences and convergence (because what is an infinite sum? It is the limit of its finite partial sums).

There is probably a "nice" way disproving the existence of such function, but I will leave this for Future-Asaf as well the other members of the site.

Answer 2

Asaf's answer explains that there is no set of all cardinals, and that under the axiom of choice, the cardinals are well-ordered, and so it is impossible to have a homomorphism as you want.

What remains is to see whether it is possible, in some models where the axiom of choice fails, to have a homomorphism $p$ as you suggest, with range a collection of sets, no two of which have the same cardinality.

This is actually possible. In fact, it holds in models where the axiom of determinacy is true, but the argument is sophisticated, and follows from recent work that started with results of Alexander Kechris and Scot Adams on Borel equivalence relations. The original reference is "Linear algebraic groups and countable Borel equivalence relations", Journal of the American Mathematical Society, Vol 13 (4), (2000) 909-943. This paper is explicitly about Borel equivalence relations, but their arguments hold in more general contexts, assuming that all sets of reals are "sufficiently nice" in ways that the axiom of determinacy guarantees.

Benjamin Miller suggested the following general statement (I believe in this specific form the result is due to him):

If $\le$ is an analytic (i.e., $\Sigma^1_1$) partial order on a Hausdorff space $X$, then there is a sequence of countable analytic equivalence relations $(E_x)_{x \in X}$ such that $$ x < y \Longrightarrow | X/E_x | < | X/ E_x \sqcup X/ E_y | = | X/ E_x \times X/E_y | = | X/ E_y | $$ for all $x, y \in X$.

Here, analytic means the continuous image of a Borel set. Without choice, an inequality $|A|<|B|$ simply means that there is an injection from $A$ into $B$ but not one from $B$ into $A$. $A\sqcup B$ is the disjoint union of $A$ and $B$; this is the canonical set associated to the cardinal sum $|A|+|B|$.

The reason why this solves the problem is that the ordering in the space $A/\sim$ is analytic in this sense, and so we can assign to each class $[a]$ the set $p([a])=X/E_{[a]}$ for $(E_{[a]})_{[a]\in A/\sim}$ a sequence of countable equivalence relations as granted by the statement above.

What follows is Ben's (very high level) sketch:

(1) In all of these results, the key is the use of what we call "rigidity arguments", which provide us with aperiodic countable Borel equivalence relations $F_x$ equipped with ergodic, invariant Borel probability measures $m_x$ with the property that for all distinct $x, y \in X$, the pair $(F_x, m_x)$ is $F_y$-ergodic (i.e., if $C$ is a set of full $m_x$-measure and $f$ is an $m_x$-measurable homomorphism from $F_x\upharpoonright C$ to $F_y$, then there is a subset $B$ of $C$ with $m_x(C \setminus B) = 0$ such that $f[B]$ is contained in a single $F_y$-class).

(2) Let $E_x$ denote the disjoint union of the equivalence relations of the form $F_y$, for $y \le x$.

(3) We have that $x \le y$ implies that the identity map is a reduction of $E_x$ to $E_y$, i.e., it gives us an injection of $X/E_x$ into $X/E_y$.

(4) If $x \le y$ is false, however, then any homomorphism from $E_x$ to $E_y$ would restrict down to a homomorphism from $F_x$ to $E_y$. Ergodicity of $m_x$ would then give a set $C$ of full $m_x$-measure and some $z \le y$ such that the map in question is a homomorphism from $F_x\upharpoonright C$ to $F_z$. As $x\ne z$, the map must therefore concentrate $m_x$-almost everywhere on a single $F_z$-class, and this implies that the original map could not have been a reduction. Thus $X/E_x$ cannot inject into $X/E_y$.

(5) From the construction one can directly argue that (for any $x$) $X/E_x \times X/E_x$ is in bijection with $X/E_x$. From this, it is straightforward to check (using Schroeder-Bernstein) that if $x\le y$ then $X/E_x\sqcup X/E_y$, $X/E_x\times X/E_y$, and $X/E_y$ are all in bijection with one another.