According to wikipedia,

...the Löwenheim–Skolem Theorem states that for every signature $σ$, every infinite $σ$-structure $M$ and every infinite cardinal number $κ ≥ |σ|$, there is a $σ$-structure $N$ such that $|N| = κ$ and

  • if $κ < |M|$ then $N$ is an elementary substructure of $M$;
  • if $κ > |M|$ then $N$ is an elementary extension of $M$.

Now let $\kappa$ be inaccessible. Then $V_\kappa$ agree with the ambient universe about $\aleph_1,$ furthermore $V_\kappa$ knows that $\aleph_1$ is uncountable. So by Lowenheim-Skolem, we can find a countable elementary substructure $M$ of $V_\kappa$. Since $V_\kappa$ is well-founded, so too is $M$. Hence, we can collapse $M$ to obtain an (isomorphic) transitive model $T$. But since $T$ is countable, hence $\aleph_1^T$ does not equal $\aleph_1.$ Thus certainly, it is not the case that the inclusion $T \hookrightarrow V_\kappa$ is an elementary embedding. Yet there exists an elementary embedding $T \hookrightarrow V$.

This seems kind weird. Can anyone explain what's going on here?

  • 1
    $\begingroup$ The elementary embedding is the isomorphism from $T$ to $M$ followed by the inclusion of $M$ in $V_{\kappa}$. I don't understand what bothers you. $\endgroup$
    – user115940
    Jun 13, 2014 at 11:53
  • $\begingroup$ It might be weird, but it is not unbelievable. Consider for example $(\mathbb N, <)$ and a substructure $(\mathbb N^+, <)$. It is not an elementary substructure, yet they are isomorphic (and hence there is an elementary embedding). $\endgroup$ Jun 13, 2014 at 12:00

1 Answer 1


This is known as the Skolem Paradox (often used with $\Bbb R$ rather than $\aleph_1$, but the principle is the same).

If you're willing to accept an inaccessible cardinal, then you are willing to accept far weirder things than "we can find other embeddings which are not inclusion".

$M$ and $T$ are isomorphic, but they are not the same. Incidentally, if you replace each element $x\in V_\kappa$ by $x\cup\{V_\kappa\}$ then you still have a natural way of identifying the resulting set with $V_\kappa$, so it is a model of $\sf ZFC$, but suddenly... the empty set is not empty.

Why is my example preposterous and the Skolem paradox isn't? Because mine is a hyperbole, of course. But the idea is the same. There is usually more than one way of doing things, especially when things are infinite.

If you want to keep insisting that something here is wrong, then you need to reject first-order logic altogether. But then you run into other problems.

  • $\begingroup$ I'm just finding it hard to imagine what's going on. Does the inclusion $T \hookrightarrow V_\kappa$ just skip most of the ordinals or something? $\endgroup$ Jun 13, 2014 at 12:44
  • $\begingroup$ What do you mean when you say "inclusion"? $T$ is transitive, so its ordinals are all countable (and there's no "skipping"). The elementary embedding is not the inclusion. It's the inverse of the collapse from $M$ to $T$. $\endgroup$
    – Asaf Karagila
    Jun 13, 2014 at 12:49
  • $\begingroup$ Ah yes sorry, that was a (fairly major) typo. I meant the elementary embedding $T \rightarrow V_\kappa$. Call it $\eta$. Now suppose $\alpha \in T$ is an ordinal. Then $\eta(\alpha)$ is an ordinal, right? And $\eta(\alpha+1) = \eta(\alpha)+1$. So we get a stretch of agreement, a long stretch of ordinals $\alpha$ where $\eta(\alpha)=\alpha$. Does there come a point in $T \cap \mathrm{ORD}$ where $\eta$ just starts growing stupendously fast or something? $\endgroup$ Jun 13, 2014 at 12:53
  • $\begingroup$ Yes, it skips many ordinals. Note that $M$ itself knows only about a small part of $V_\kappa$, so $T$ doesn't "reach all the way up" in this embedding. $\endgroup$
    – Asaf Karagila
    Jun 13, 2014 at 13:02
  • $\begingroup$ Okay, I think I'm starting to get it now. Since $\aleph_1$ is regular, hence there must be an ordinal $\alpha<\aleph_1$ such that no $\beta \in [\alpha,\aleph_1)$ is an element of the image of $\eta.$ Otherwise, we run into cofinality problems. Right? $\endgroup$ Jun 13, 2014 at 13:22

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