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My book says

"... If $GCH$ holds, then the notions of strongly inaccessible and weakly inaccessible cardinals coincide, ..."

In $ZFC$ I can prove this. But the paragraph from which I have excerpted this sentence starts with

"... Apparently we have found a set model of $ZF$. ..."

Which suggests that perhaps we have strongly = weakly inaccessible in $ZF + GCH$.

Can one show in $ZF + GCH$ that weakly inaccessible implies strongly inaccessible?

My definition of choice of cardinals in the absence of choice is $|A| = \{ B : B \approx A \text{ and } B \in V_\beta \}$ where $\beta$ is the smallest ordinal such that there exists a $B$ in $V_\beta$ that is in bijection with $A$.

Many thanks for your help.

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I would assume that in choiceless contexts GCH should be taken to say that for any $X$, $|\mathcal{P}(X)|$ is the smallest cardinality of a set into which $X$ injects and doesn't biject. – Miha Habič Dec 21 '12 at 21:11
@MihaHabič I upvoted your comment but then I realised that it is not clear to me why you would want to modify $GCH$. We still have $\aleph$ numbers without $AC$ so the statement $GCH$ still makes sense without $AC$. – Rudy the Reindeer Dec 21 '12 at 21:15
That's true, but I imagine that without choice you can have nonwellorderable sets whose cardinalities behave like regular limit cardinals. GCH in its standard form wouldn't tell us anything about these. I guess this all depends on what you mean by cardinal in a choiceless world. – Miha Habič Dec 21 '12 at 21:32
GCH implies choice, even if you only assume GCH for initial ordinals. – Andrés E. Caicedo Dec 21 '12 at 22:25
Dear @AndresCaicedo, thank you, your comment answers my question. – Rudy the Reindeer Dec 22 '12 at 7:34
up vote 3 down vote accepted

First I should point out that there is some correction to your definition of a cardinal. If there is an ordinal which is in bijection with $A$ then $|A|$ is the least such ordinal; otherwise it is that set which you described.

Now I should point out that there are two ways to formulate GCH when the axiom of choice fails.

  1. For every ordinal $\alpha$, $2^{\aleph_\alpha}=\aleph_{\alpha+1}$. This is known in some places as the Aleph Hypothesis.
  2. For every set $X$ if $Y$ is such that $|X|\leq|Y|<|\mathcal P(X)|$ then $|Y|=|X|$. This is known as the Generalized Continuum Hypothesis.

Clearly under the axiom of choice both are equivalent. As it turns out, however, both of these are actually equivalent without the axiom of choice because both of these imply that the axiom of choice.

If ZF+GCH imply the axiom of choice then your question is trivially yes, because as you remarked the implication in ZFC is simple.

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