The Löwenheim–Skolem theorem shows that we can find a countable elementary submodel of $V$ that satisfies $ZFC$. [assuming, Con$(ZFC$)]. Call this set $U$. Then by the definition of elementary submodel, $V$ and $U$ must believe the same formulae. Let $\kappa$ be a cardinal in $U$ that $U$ believes to be uncountable. (Such a cardinal must exist as $V$ believes that there are uncountable cardinals, therefore so does $U$). Then as $U$ countable, $\kappa$ must be countable (as seen from $V$). However, now $U$ and $V$ disagree about the formula '$\kappa$ is uncountable', which seems (to me) to contradict the definition of elementary submodel. Where have I gone wrong here?
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$\begingroup$ Note that Lowenheim-Skolem only applies to sets, not classes. But assuming the existence of an inaccessible cardinal $\lambda$, you could replace $V$ with $V_\lambda$ in your argument. $\endgroup$– Eric WofseyCommented Feb 8, 2018 at 23:25
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$\begingroup$ @EricWofsey Or work in an appropriate class theory. $\endgroup$– Noah SchweberCommented Feb 8, 2018 at 23:27
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1$\begingroup$ (Or use the reflection principle to get a $U$ which agrees with $V$ on the finite collection of formulas you need for the argument.) $\endgroup$– Eric WofseyCommented Feb 8, 2018 at 23:28
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$\begingroup$ @EricWofsey Yes, I think that's the best thing to do here, since it doesn't require a jump in consistency strength. $\endgroup$– Noah SchweberCommented Feb 8, 2018 at 23:30
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1$\begingroup$ Words to live by: "Skolem's Paradox". $\endgroup$– Asaf Karagila ♦Commented Feb 9, 2018 at 8:56
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Good question! This is a subtle point. The error is when you write:
$(*)$ Then as $U$ countable, $\kappa$ must be countable (as seen from $V$).
This is not the case! Presumably, the reason for believing $(*)$ is (something like) "$\kappa$ in $U$, so $\kappa\subseteq U$," but this assumes that $U$ is transitive. (A set $A$ is transitive if $y\in x\in A\implies y\in A$.)
This need not be the case; in fact, your exact argument shows that $U$ is never transitive! Rather, all we can conclude from the countability of $U$ is that the set $$\kappa\cap U$$ must be countable (as seen from $V$). Basically, $U$ will contain lots of elements which are uncountable sets, but $U$ will contain only "countably much" of each.
answered Feb 8, 2018 at 23:21
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1$\begingroup$ @ElieBergman All first-order formulae are indeed absolute between $U$ and $V$, in the following sense: if $\overline{a}\in U$, $\varphi$ is a first-order formula, and $U\models\varphi(\overline{a})$, then $V\models\varphi(\overline{a})$. What do you think is a non-absolute formula? $\endgroup$ Commented Feb 8, 2018 at 23:43
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1$\begingroup$ (A possibly important point here: note that $\kappa\cap U$ will not, in general, be an element of $U$, even if $\kappa$ is! We have to be very careful when thinking about what $U$ does and doesn't "see.") $\endgroup$ Commented Feb 8, 2018 at 23:46
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1$\begingroup$ @ElieBergman "$\kappa$ is uncountable" is true in both models. Why do you think it isn't? The set which is countable is $\kappa\cap U$, but as I remarked above $\kappa\cap U$ isn't even an element of $U$. $\endgroup$ Commented Feb 9, 2018 at 11:49
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1$\begingroup$ @ElieBergman By the way, here's a neat fact about elementary submodels. Suppose $M\prec V$. Then $\omega_1\in M$ (since $\omega_1$ is definable) but in general $\omega_1\not\subseteq M$ (e.g. if $M$ is countable). Now for a lot of arguments, the set "$\omega_1\cap M$" plays an important role. At first one might be worried that this set could be extremely complicated, but it turns out that it's actually rather nice - $\omega_1\cap M$ is always closed downwards! That is, if $\alpha$ is a countable ordinal in $M$ and $\beta<\alpha$ then $\beta\in M$. Alternatively, $\omega_1\cap M$ is an ordinal. $\endgroup$ Commented Feb 10, 2018 at 20:37
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1$\begingroup$ This is a special feature of $\omega_1$: for example, $\omega_2$ is also definable, hence $\omega_2\in M$, but if $M$ is countable we won't have $\omega_2\cap M$ closed downwards (since we'll have $\omega_1\in M$ but $\omega_1\not\subseteq M$). The proof of the fact above is quite neat. Suppose $\alpha\in M$ is a countable ordinal. By elementarity, $M$ contains a function $f$ which $M$ thinks is a bijection from $\alpha$ to $M$. By appropriate absoluteness, $f$ is a bijection from $\alpha$ to $\omega$ in $V$. Finally, note that $\omega\subseteq M$ since each finite ordinal is definable. $\endgroup$ Commented Feb 10, 2018 at 20:39