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In Just Weese on page 197 there are the following corollaries:

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Regarding Corollary 24: Is this a typo and should say "$CON(ZF) \not\rightarrow CON(ZF + \exists \text{ "a strongly inaccessible cardinal."})$"?

I can't see anything wrong with the implication "$CON(ZF) \rightarrow CON(ZF + \text{ "there are no strongly inaccessible cardinals."})$".

Many thanks for your help.

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    It does seem like a typo.2012-12-22
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    @AsafKaragila What a typo to make : / Fatal. I think the authors got bored of writing towards the end of the book.2012-12-22
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    So they figured "Hey, let's add particularly misleading typos and see if anyone is reading the book carefully enough at this point!"2012-12-22
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    @AsafKaragila Can you have a look at page 199 and confirm that there is yet another fatal typo? On page 199 there is a paragrpah starting with "How can one construct..." and in it we have $\mathcal P^{\mathbf X}(x) \subseteq \mathcal P^{\mathbf X}(x)$. I think this should be $\mathcal P^{\mathbf V}(x) \subseteq \mathcal P^{\mathbf X}(x)$. And similarly, in the sentence following this one the second occurrence of $\mathcal P^{\mathbf V}(x)$ should be $\mathcal P^{\mathbf X}(x)$.2012-12-22
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    It should be $\mathcal P^\mathbf X(x)\subseteq\mathcal P^\mathbf V(x)$, I think. In the second line as well. We want $\bf X\subseteq V$, so it only makes sense that the power set in $\bf X$ is not-larger or even smaller.2012-12-22

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The next two paragraph give a proof of Corollary 24. The content of this proof is that if there are no inaccessible cardinals then we are done; otherwise there is a least inaccessible $\lambda$, in which case Lemma 25 proves that $\langle V_\lambda,\overline\in\rangle$ is a model of ZF+"There are no inaccessible cardinals".

In either case we see that the proof actually proves:

$CON(\mathrm{ZF})\nrightarrow CON(\mathrm{ZF}+\exists\text{ a strongly inaccessible cardinal})$

Which also fits the two-liner proof based on Godel's incompleteness theorem which is given right after Corollary 24.

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    And there is nothing wrong with the other implication, right?2012-12-22
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    @Matt: The proof given actually shows that the implication holds, so indeed there is nothing wrong with it.2012-12-22