$H(\kappa)=\{x:|TC(x)|<\kappa\}$, with $\kappa$ regular card. and $TC(*)$ the transitive closure.
Now we defined an elementary substructure:
$M\prec H(\kappa)$ if for every formula $\varphi$ and all $a_{1},\dots, a_{n}\in M$ holds: $\varphi^{M}(a_{1},\dots,a_{n})\leftrightarrow\varphi^{H(\kappa)}(a_{1}.\dots, a_{n})$.

My first question is, what does this exactly mean? It was the end of the lecture, and so we defined this quite hastily.

And afterwards the professor wrote:
Let $\kappa > \omega_{1}$ and $M\prec H(\kappa)$ be a countable elementary substructure.
Note: $M$ is defenitely not transitive.
How do I see that fast, that $M$ can not be transisive?
(Definition: set $T$ transitive, if $x\in T$ implies $x\subset T$.)

  • 4
    $\begingroup$ If $\kappa>\omega_1$ then $\omega_1$ is definable in $H(\kappa)$ as the least ordinal such that no function from $\omega$ onto that ordinal exists (notice that any such function's transitive closure is countable). Then $M$ has an ordinal like this, and this ordinal has to be $\omega_1$, because otherwise (if it was $\alpha<\omega_1$) if $\phi(x)$ is the formula "$x$ is least such that no function exists from $\omega$ onto $x$", then $M\models\phi(\alpha)$ while $H(\kappa)\nvDash\phi(\alpha)$. So $\omega_1\in M$ but since $M$ is countable, $\omega_1\nsubseteq M$. $\endgroup$ – Apostolos May 28 '13 at 18:58
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    $\begingroup$ Allow me to confuse you even more at this point, even though $M$ is not transitive, it is well-founded, and therefore can be collapsed to a countable transitive structure, $N$. It follows that there is some countable ordinal $\alpha$ such that $N\models|\alpha|>\aleph_0$. Strange, isn't it? :-) $\endgroup$ – Asaf Karagila May 28 '13 at 19:13
  • $\begingroup$ (Pedagogically I am not usually in favor of confusing students, but this time I'm making an exception, because understanding this can save a lot of trouble later on) $\endgroup$ – Asaf Karagila May 28 '13 at 19:14
  • $\begingroup$ Thanks all of you for the explanation! @Asaf, could you maybe say some more words to this? How does it follow out of $M$ being well-founded? I can also put it as a new question if it gets too chaotically here. $\endgroup$ – Luca May 28 '13 at 19:35
  • 3
    $\begingroup$ Set theorists often talk about collapsing classes which are well-founded. They're not trying to be Marxists or something like that, they just refer to the Mostowski collapse lemma. $\endgroup$ – Asaf Karagila May 28 '13 at 20:20
  1. This means that the same first order sentences are true, using one binary relation symbol in the language, interpreted as '$\in$'.

    $M$ is a substructure of some $(T,\in)$ means that the relation symbol of $M$ is also interpreted as $\in$, and that it is an elementary substructure means that in addition all formulas have the same truth values when variables are evaluated in $M$.

  2. $\kappa>\omega_1$ implies that $H(\kappa)\models\,$"$ \exists x:\nexists f:x\hookrightarrow \omega$".
    This latter one means that there is a set which cannot be embedded into $\omega$, and this sentence can be written as a correct first order formula, using only '$\in$'.

    By 1., the same sentence also holds in $M$, though it doesn't hold in any countable transitive set.


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