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The existence of a standard model of ZF is stronger than the assertion of its consistency, but this is not what I mean.

It is impossible to first-order say that there is a standard model of ZF. It goes without saying that every model of ZF+Con(ZF) or ZF+"there is a standard model" could be nonstandard.

When doing consistency proofs with forcing (say, an internally countable forcing notion), it is meaningless to try to extend a countable nonstandard model $M$ to have a standard-countable dense set. But these forcing independence results must follow strictly from ZF. It seems that for this type of forcing we can remedy this by re-indexing our countable dense subset according to the possibly nonstandard $\mathbb{N}^M$, but point notwithstanding (in light of internally-uncountable forcing notions or the subtleties of class forcing, etc):

Always working in a standard model of ZF or its extensions seems awfully reckless. How do I come to accept that we can always do this without giving it a second thought?

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    Is CH true in the standard model ? – 2017-02-05
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    @ReneSchipperus No. From a standard model of ZFC we can generate a standard model of $\mathsf{ZFC + \lnot CH}$ via forcing. – 2017-02-06
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    Having stood in your shoes before, my advice is to put these issues aside for now. You understand them better when you understand how forcing works, how independence is proved, and the myriad of small, but significant, issues between theory and metatheory. – 2017-02-06
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    @AsafKaragila Yeah what do they mean by standard model of ZFC ? – 2017-02-06
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    @Rene: Transitive model (or just well-founded model). This is, err, standard terminology. – 2017-02-06
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    @AsafKaragila Can one show that if $M$ is a, say countable, model of $ZFC+\varphi$ then there is a countable transitive model of the same ? – 2017-02-07
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    @Rene: No, since standard models must agree with the universe on arithmetic statements, taking $\varphi$ to be something like "ZFC is not consistent", then it cannot hold in a standard model. – 2017-02-07
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    @AsafKaragila Yes of course a transitive model is absolute for arithmetic sentences. Does it go further $\Sigma_1$ ? – 2017-02-07
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    @Rene: Not sure what you mean by that. But I feel that I don't know the answer either. – 2017-02-07
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    @AsafKaragila Isnt there a hierarchy of formulas above the arithmetic ? Quantification over subsets of $\mathbf{N}$. Isnt $L$ $\Sigma_1^1$ absolute or something like that ? – 2017-02-07
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    @Rene: Yes, and Shoenfield's absoluteness tells us that $L$ is $\Pi^1_2$-absolute. I don't know the answer though. (And I wasn't sure if you're talking about $\Sigma_1$ in the Levy hierarchy or $\Sigma_1^1$ or $\Sigma_1^0$ in the projective and arithmetical hierarchies.) – 2017-02-07
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    @AsafKaragila Yeah, see how poor my memory is, Ill have to review my set theory, so I guess my question was really is there "some" version of Shoenfield valid for transitive models, maybe with weaker than $\Pi_2^1$ ? – 2017-02-07
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    @Rene: Mostowski has a weaker version, yes. Also, drop me an email. I need to tell you something unrelated. – 2017-02-07

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