In the book 'A Mathematical Introduction to Logic' by Enderton, he stated the Skolem's paradox at page $152:$

Let $A_{ST}$ be your favourite set of axioms for set theory. We certainly hope these axioms are consistent. And so they have some model. By Lowenheim-Skolem Theorem, they have a countable model $\mathfrak{S}.$ Of course, $\mathfrak{S}$ is also a model of all the sentences logically implied by $A_{ST}.$ One of these sentences asserts that there are uncountable many sets.

There is no contradiction here, but the situation here is sufficiently puzzling to be one called 'Skolem's Paradox'. It is true that in the structure $\mathfrak{S}$ there is no point that satisfies the formal definition of being a one-to-one map of the natural numbers onto the universe. But this in no way excludes the possibility of there being (outside $\mathfrak{S}$) some genuine function providing such a one-to-one correspondence.

I fail to understand the bold sentences.

From what I read online, the fact that $\mathfrak{S}$ is countable is due to external viewing of $\mathfrak{S},$ that is, if we view universe of $\mathfrak{S}$ as a sub-universe, then we can find a bijection $f$ in the bigger universe between $\mathfrak{S}$ and set of natural numbers, and hence $\mathfrak{S}$ is countable. By the way, $\mathfrak{S}$ does not know that it is countable. (WHY?)

Cantor's Theorem in Set Theory asserts the existence of uncountable set. So $\mathfrak{S}$ contains an uncountable set. However, being uncountable in $\mathfrak{S}$ does not mean it is uncountable in a bigger universe than universe of $\mathfrak{S},$ as there might be a bijection $g$ from a bigger universe between the uncountable set and set of natural numbers.

Is my thinking above correct? I still fail to grasp the essence of Skolem's Paradox not being a paradox.

  • $\begingroup$ To be countable according to the theory means that the theory proves the existence of a bijection. If the theory asserts that a certain set is uncountable, this means that the theory does not prove the existence of the necessary bijection. $\endgroup$ – Mauro ALLEGRANZA Mar 3 '18 at 11:20
  • $\begingroup$ @MauroALLEGRANZA: But how do we guarantee that the theory does not prove the existence of bijection? I am confused. $\endgroup$ – Idonknow Mar 3 '18 at 11:22
  • $\begingroup$ The theory $T$ proves Cantor's Th, asserting that there is a set (call it $A$) which is uncountable. Assuming that the theory is consistent, if the theory proves also that there is a bijection $f: A \to \mathbb N$, we have found a contradiction. Thus, if $T$ is consistent, it does not prove the existence of $f$. $\endgroup$ – Mauro ALLEGRANZA Mar 3 '18 at 11:48
  • $\begingroup$ But the topis has many interesting (and some controversial) issues; see e.g. Skolem's Paradox. $\endgroup$ – Mauro ALLEGRANZA Mar 3 '18 at 11:50
  • $\begingroup$ Assuming that $T$ is consistent, it has a model: $\mathfrak{S}$. All theorems of $T$ are true in it; thus, in it there is a set $A$ such that the theory does not "know" of any bijection in $\mathfrak{S}$. $\endgroup$ – Mauro ALLEGRANZA Mar 3 '18 at 11:56

Simplyfying, the $\Bbb N$-in-your-structure and the $2^{\Bbb N}$-in-your-structure are countable in the "external" universe. I.e., there is a bijection $f:\Bbb N\hbox{-in-your-structure}\longrightarrow 2^{\Bbb N}\hbox{-in-your-structure}$, but your structure "ignores" the existence of $f$ because $f$ isn't a member of your structure.

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