I heard recently that for a given cardinal $\lambda$, the fact that there is a least cardinal $\lambda^+$ greater than $\lambda$ is independent of ZF... Is this true? And if so, is it possible to show why?
2026-03-25 19:11:42.1774465902
Successor cardinals independent of ZF
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This depends on what you mean by 'cardinal'. If 'cardinal' means 'initial ordinal', the answer is no:
Let $X$ be any set (e.g. $X$ can be a cardinal). Hartogs' number of $X$, given by $$ \aleph(X) := \{ \beta \in \mathrm{Ord} \mid \text{ there is some injection } \beta \to X \} $$ is the least ordinal which cannot be embedded into $X$. If $X$ itself was an initial ordinal (a cardinal), then $\aleph(X)$ is precisely the cardinal successor of $X$.
On the other hand, if you allow for non-wellorderable cardinals $X$, it's not obvious what its cardinal successor would even mean. (You could reasonably argue that it should be the Hartogs' number of $X$ -- which always exists but maybe you have something else in mind).