# Equi-consistency of inaccessible cardinals [duplicate]

I have trouble understanding this wiki page, concerning inaccessible cardinals.

First "These statements are strong enough to imply the consistency of ZFC". I guess what they mean by that is if $\kappa$ is a strongly inaccessible cardinal, then $V_\kappa$ is a model of ZFC, where $V$ is the Von Neumann hierarchy.

Then they argue that a proof in ZFC, assuming ZFC is consistent, of the consistency of ZFC+$\kappa$ is impossible. Because it would prove the consistency of ZFC, violating Gödel's incompleteness theorem.

I don't understand this, I imagine a relative consistency proof would rather be a model transformation : given a model of ZFC as input, produce a model of ZFC+$\kappa$ as output. Then we could continue producing $V_\kappa$ to get another model of ZFC. Where is the contradiction in that ? To me it seems we first assumed a model of ZFC, then produced another one. ZFC didn't produce a model of itself from scratch.

## marked as duplicate by Asaf Karagila♦ set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 4 '18 at 15:43

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## 1 Answer

Apply the incompleteness theorem to $ZFC+\exists \text{an inaccessible}$

• So simple... sorry I didn't see that myself – V. Semeria Aug 4 '18 at 15:58