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I do not understand the following passage/footnote in the book I am currently reading:

An initial ordinal $\lambda$ is called a strong limit cardinal if $2^\kappa < \lambda$ for every $\kappa < \lambda$. Note that this definition makes sense even in $ZF$.

Then, footnote:

If $\lambda$ is a strong limit cardinal, then $\mathcal P (\kappa) $ can be wellordered for every $\kappa < \lambda$. Thus, in $ZF$ alone one cannot prove the existence of an uncountable strong limit cardinal. But it is also relatively consistent with $ZF$ that an uncountable strong limit cardinal exists while the full Axiom of Choice fails.


What I understand: $\lambda$ is a cardinal, hence in particular an ordinal and hence well-ordered. If $2^\kappa < \lambda$ we therefore get a well-order on $\mathcal P (\kappa)$ induced by an order-isomorphism between $\mathcal P (\kappa)$ and some ordinal $\alpha < \lambda$.

What I don't understand: The sentence starting with "Thus...". How does it follow from $\mathcal P (\kappa)$ being well-ordered that in $ZF$ one cannot prove the existence of an uncountable strong limit cardinal? And: Would you show me how to prove the existence of an uncountable strong limit cardinal?

Many thanks for your help.

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    You need choice to say that every cardinal is an ordinal. If every cardinal is an ordinal then every set $x$ can be well-ordered by the order induced by the bijection $x \to |x|$.2012-12-21

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