# Assuming $(GCH)$, strongly inaccessible and weakly inaccessible coincide

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 $A$.

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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. –  Andres 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

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.
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.