# Standard models being non-standard?

If there is a ''set'' W in V which is a standard model of ZF, and the ordinal κ is the set of ordinals which occur in W, then Lκ is the L of W. If there is a set which is a standard model of ZF, then the smallest such set is such a Lκ. This set is called the minimal model of ZFC. Using the downward Löwenheim–Skolem theorem, one can show that the minimal model (if it exists) is a countable set.

Of course, any consistent theory must have a model, so even within the minimal model of set theory there are sets which are models of ZF (assuming ZF is consistent). However, those set models are non-standard. In particular, they do not use the normal element relation and they are not well founded. (http://en.wikipedia.org/wiki/Constructible_universe)

I don't get it. OK, so in this case, one assumes that there exists some standard model that is not a class, but a set. Then one constructs a constructible universe inside such set which is also a set. Then one uses Lowenheim-Skolem to show that the smallest such model is countable.

My question is, it seems that the model is externally not standard (viewed from a bigger model) - so why call it standard first place? Or am I being confused, and is the model really standard? And the quote in the later part seems to say that the model is not standard, which confuses me much.

And I do not get the second paragraph - why would any consistent theory having at least one model relates to the minimal model containing models of ZF? I am not sure how Godel's completeness theorem is being matched here.

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If you have a standard model $M$, and $N$ is a countable submodel (not even elementary, just a substructure which is a model of $\sf ZFC$) then $N$ is well-founded, because its $\in$ is a subset of the real $\in$ of the universe of sets (which is well-founded by the axiom of foundation).
We can therefore collapse $N$ to a countable transitive model, which is then a standard model as wanted.
Within the minimal model, however, we can't do this sort of trick anymore. There is no set which is a standard model that we can begin with. However the standard integers cannot encode a proof of $\lnot\text{Con}(\sf ZFC)$, so the minimal standard model also satisfies $\text{Con}(\sf ZFC)$ and therefore has a model. But this model cannot be standard, it has to be ill-founded.
@Steven: I'm not 100% sure about your request (but I'll be happy to add more once I am sure about it). Models of set theory may not be aware that there is a decreasing sequence $f_{n+1}\mathrel{E} f_n$ such that $f_n\in M$ and $E=\in^M$. So even if every $f_n$ is in the model, the set $\{f_n\mid n\in\omega\}\notin M$ -- so the model does not know about this violation of the axiom of foundation. – Asaf Karagila Jun 26 '13 at 15:41
I think that's just what I was after: while there is such a sequence (and so 'externally' the model is ill-founded) the internal model itself isn't aware of it because it can't 'collect' all the $f_i$. That makes sense to me at least. – Steven Stadnicki Jun 26 '13 at 15:48
And thinking about it more: this doesn't violate $\omega$-consistency either, I presume, because it's not a question of having some non-exhibited example, but simply of not having the collection of the $f$ - which really makes it almost exactly identical to the situation of internal-vs.-external countability in countable models. – Steven Stadnicki Jun 26 '13 at 17:00