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I am curious about why the existence of inaccessible cardinals cannot be proven in ZFC.My intuitive proof is:

Suppose we can prove there is an inaccessible cardinal ϰ,then by definition we must show that for any cardinals b&c which satisfying b<ϰ and c<ϰ,then b^c<ϰ.Then if we consider the set S which is the union of all the cardinals less than ϰ and consider the power set of S--P(S).It follows that |P(S)|>=ϰ.This implies we can proof the existence of inaccessible cardinals under ZFC iff we can disprove it.Therefore we cannot prove the existence using ZFC.

Is this the intuitive idea behind the proof?

I only have learnt a little bit in set theory by self-studying.But I don't know any thing about mathematical logic.So I don't know the thing such as the second incompleteness theorem.This statement was found in my set theory book but without prrof it.So any comment and improvement of my work is welcome!

Asaf Karagila
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Andrew
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  • The way you argue you actually show that inaccessibles are inconsistent with ZFC, which is more than just saying their existence cannot be proven. But anyway your argument is false as pointed out in an answer. – Jonathan Schilhan Jan 08 '17 at 10:49

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The simplest proof for why we cannot prove the existence of inaccessible cardinals is as follows:

If $\kappa$ is inaccessible, then $V_\kappa$ is a model of $\sf ZFC$. Now if the existence of inaccessible cardinals was provable, look inside $V_\kappa$, where $\kappa$ is the least inaccessible cardinal. There you should find some $\lambda$ which is inaccessible as well. However this would imply $\lambda$ is also inaccessible, which is a contradiction since $\kappa$ was the minimal.

Asaf Karagila
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  • Asaf Karagila:So your goal is you want to show that if the first inaccessible cardinal ϰ is constructible by using cumulative hierarchy even with all the axioms in ZFC in λ steps where λ<ϰ.So λ must be an inaccessible cardinal by definition.Therefore we reached a contradiction.In conclusion,the core of the proof is we want to show that ϰ is ''too large'' such that it isn't constructible in the range of ZFC.Is this what you want to say? – Andrew Jan 08 '17 at 16:01
  • To the OP; If $k$ is strongly inaccessible and $x<k$ then ($x$ is strongly inaccessible)$\iff (V_k,\epsilon) \vDash (x$ is strongly inaccessible). – DanielWainfleet Jan 08 '17 at 18:05
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    @Andrew: I'm not sure what you're trying to say. We don't "construct" $\kappa$ using the cumulative hierarchy. The cumulative hierarchy is given with the universe. If $\kappa$ is the smallest inaccessible, then the construction of the hierarchy up to $\kappa$ is a model of ZFC; so by the assumption there is some $\lambda$ there which is inaccessible. But being inaccessible inside $V_\kappa$ is the same as being inaccessible in the whole universe. Therefore $\kappa$ wasn't the smallest inaccessible. – Asaf Karagila Jan 08 '17 at 18:14
  • Asaf Karagila:Sorry but I still cannot understand how can you therefore conclude that there is some inaccessible cardinals λ inside Vϰ? – Andrew Jan 08 '17 at 18:38
  • @Noah: The OP said they are not familiar with the incompleteness theorem. So I rather not appeal to that. – Asaf Karagila Jan 08 '17 at 19:07
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    @Andrew: Since $V_\kappa$ is a model of ZFC, if ZFC proves there is an inaccessible cardinal, then in $V_\kappa$ there is some $\lambda$ such that $V_\kappa\models\lambda\text{ is inaccessible}$. – Asaf Karagila Jan 08 '17 at 19:08
  • @AsafKaragila Quite right, sorry. Still, you do need to argue that if $V_\kappa\models$"$\lambda$ is inaccessible," then $\lambda$ is actually inaccessible. – Noah Schweber Jan 08 '17 at 19:15
  • @Noah: Yes, but that does not require you sticking your hands into the logic of things. Just the fact that $V_\kappa$ is correct about power sets. – Asaf Karagila Jan 08 '17 at 19:45
  • @AsafKaragila True - my point was that the OP might not see that immediately, so it's worth pointing out. – Noah Schweber Jan 08 '17 at 19:56
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You proof fails because because by definition if $\kappa$ is inaccesible then is not a sum of fewer than $\kappa$ cardinals that are less than $\kappa$. For constructing $S$ you are adding $\kappa$ cardinals.

For the usual proof see Proving that the existence of strongly inaccessible cardinals is independent from ZFC?.

Community
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Martín-Blas Pérez Pinilla
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If we could prove that a strongly inaccessinle $k$ exists from $ZFC$ we could then prove $Con [ZFC]$ because the set $(V_k,\epsilon)$ would a model for $[ZFC].$ By Godel's First Incompleteness Theorem, if $Con[ZFC]$ is a theorem of ZFC then ZFC is inconsistent. If ZFC really is inconsistent then we're all wasting our time. If it's consistent, it can't prove such $k$ exist.

$[ZFC]$ is a numerical encoding of ZFC. In the language of set theory we cannot make statements about sentences of the language but we can talk about numbers (E.g. ASCII codes for strings of keyboard characters) and relations between them.

DanielWainfleet
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