Skolem's paradox showing us that we might be trapped in our view of the world

Question

According to Skolem's Paradox, ZFC as a first order axiomatization of set theory has a countable model, but allows a proof that uncountable sets exist in every model of ZFC.

It becomes counter-intuitive if we accept ZFC as modelling our reality, a formal model of what we think a set is, but also accepting Cantor's proof that $\mathbb R$ is uncountable.

The paradox could be resolved by looking at the notion of "uncountability" within the axiom system (and applied to a particular model), and externally (a notion we have of uncountable outside of the axiom system). This is the explanation from nlab, to cite:

The resolution of this apparent paradox is that, while this conclusion is true internally, it is not true externally: namely any two infinite sets are countable externally in that model, hence there is a $1$–$1$ function between any two of them including for a model of some uncountable set $X$ and of its power set $P(X)$. However, that function (or its graph) is not in the model! One can enlarge the model by adding the function (and more). But this extended model will necessary have $P(X)$ uncountable externally and there is no $1$–$1$ function from $X$ to $P(X)$ externally any more.

Now assume that ZFC captures what we think of as a set. The real numbers seem perfectly uncountable for us, and somehow they capture the idea of approximation on the number line. Would it be possible that another observer with "more insight" looks at us, at our model of ZFC, and for them this would look countable, as they are able "externally" to put our construction of $\mathbb R$ in correspondence from our construction of $\mathbb N$.

I would say no, simply because I cannot imagine, simply because I think I believe in the construction. But Skolem says that we cannot be sure about our interpretation of ZFC. But what do you think, may we be trapped in some model of thinking? Is this a conceivable scenario, another observer being smarter than us, seeing our reasoning "externally"? But then this seems to be a new incarnation of Skolem's paradox right from its resolution, with giving us some sort of epistemological boundary...

Answer 1

The multiverse view somehow got lost in the shuffle after being briefly mentioned by Henning Makholm but it is highly relevant to the OP's question. In fact, the relativity implied is even stranger than what you described in your question. Thus, Edward Nelson proved theorem 1.2 on page 1167 in

Nelson, Edward Internal set theory: a new approach to nonstandard analysis. Bull. Amer. Math. Soc. 83 (1977), no. 6, 1165–1198.

Nelson's theorem implies that there exists a finite set $F$ such that all standard real numbers are contained in $F$.

Now the "baby model" of Hamkins' multiverse can be viewed as providing precisely the nested sequences if worlds that the OP envisioned, each one being the standard objects of the larger one in such a way that the axioms of a set theory (BST) closely related to Nelson's IST is satisfied. This is discussed in more detail in this 2017 publication in Real Analysis Exchange.

Thus, all standard reals of a subworld actually belong not merely to a countable but to a finite set in the superworld in this scheme.

See also this related post.

Answer 2

If understand correctly you are assuming that there is some sort of real model of sets, that is a collection of things we intuitively understand as sets and an intuitive membership relation between such things. These data satisfy ZFC axioms.

Now you said:

Now assume that ZFC captures what we think of is a set.

By that I assume you mean that ZFC is the ultimate theory which should be able to characterize completely what sets are. If my interpretation of this statement is correct than you are wrong.

First of all by Rosser's theorem (one of the generalization of Gödel's incompleteness theorem) we know that (if consistent) ZFC is incomplete, meaning that there are formulas for sets that cannot be proven nor refused by ZFC. In particular any such formula should either be true or false in the real model of sets but the problem is that ZFC does not allow to tell which case is the correct one.

So ZFC does not really capture what a set is (if by set we mean an object of this intuitive model).

Anyway clearly this is not the only problem. If $(\mathbb S,\in)$ is such real model of sets we could consider the first-order theory $T(\mathbb S)$ of all first-order formulas true in this real model.

Skolem's theorem would equally apply to this theory and so we could build inside $\mathbb S$ a model of $T(\mathbb S)$ which is countable, i.e. this model can be put in bijection with the set $\mathbb N \in \mathbb S$.

This means that even $T(\mathbb S)$ is not sufficient to characterize the real model of sets. But that should not be a surprise because the two models of $T(\mathbb S)$ can be distinguished by higher-order properties.

Indeed if we take the higher-order theory $T^h(\mathbb S)$ of the all the formulas in the higher-order language for $\mathbb S$ true in $\mathbb S$, and we assume we use higher-order logic with Taskian higher-order semantics, then we get a theory where there cannot be a countable model of $T^h(\mathbb S)$.

That is due to the fact that for higher-order logic (with Tarskian-semantics) there is no compactness theorem, hence no Skolem's theorem.

A similar phenomena can be observed for Peano Axioms: PA1(=first-order Peano axioms) is subject to Skolem's theorem, hence there are many models of these axioms of all the cardinalities, nevertheless PA2(=second-order Peano Axioms) has a unique model.

So, if I understood you correctly, your problems were due a wrong assumption, namely that ZFC could capture everything about sets.

I assume that the discussion above proved you otherwise, in case of any doubts or if in need of clarification please feel free to post in the comments below.

Answer 3

The "paradox" in your first sentence is not a paradox so it does not to be resolved. It is a mis-understanding due to the slipperiness of English grammar. There is no theorem that every model $M$ of $ZFC$ contains an uncountable member. If $M$ is a model of $ZFC$ then ($\exists x: x $ is uncountable$)^M.$ What this means is that for some $x\in M$ there is no $f\in M$ such that $(f:x\to \Bbb N$ is an injective function$)^M.$ But ($x$ is uncountable$)^M$ is not necessarily equivalent to ($x$ is uncountable), and if $M$ is countable then it certainly isn't.