I am having trouble gaining a basic understanding of inaccessible numbers. I've written a list of basic properties and was wondering if anyone can confirm for me if it is accurate?

$\kappa$ is inaccessible:

  • $\kappa$ is regular
  • $\alpha < \kappa \rightarrow 2^\alpha < \kappa$
  • $\kappa$ is a fixed point on the Aleph function
  • $\kappa = \aleph_{\kappa}$
  • $\kappa$ (as an ordinal) is a limit ordinal
  • $\kappa$ cannot be expressed as a sum of fewer than $\kappa$ cardinals with cardinality less than $\kappa$
  • $\kappa$ is not a successor cardinal
  • $cf(\kappa) = \kappa$

Thank you in advance!

  • $\begingroup$ The first, sixth, and last are the same; the third and the fourth are the same. $\endgroup$
    – Asaf Karagila
    Commented Jan 9, 2017 at 4:12

1 Answer 1


There are two kinds of inaccessible cardinals, weakly inaccessibles and strongly inaccessibles.

  • $\kappa$ is weakly inaccessible if it is a regular limit cardinal.

  • $\kappa$ is strongly inaccessible if it is a regular strong limit cardinal, that is, if $\kappa$ is weakly inaccessible and $2^\alpha<\kappa$ for all $\alpha<\kappa$.

Assuming the Generalized Continuum Hypothesis, these are the same thing; but that's not provable in ZFC alone.

I assume that by "inaccessible" you mean "strongly inaccessible" (I believe that is how that term tends to be used now, although older texts often use it to mean "weakly inaccessible"). In that case, everything you've written is indeed a property of inaccessible cardinals.

  • Properties 1 and 2 - together with "being a cardinal" - are the definition of inaccessible cardinal.

  • 8 is the same as 1 and 6, and 3 is the same as 4.

  • To see that $\kappa$ must not be a successor cardinal (point 7), suppose $\kappa=\lambda^+$; what does this tell you about $\kappa$ versus $2^\lambda$? Why does that contradict inaccessibility?

  • Meanwhile, 5 follows immediately from the fact that inaccessible cardinals are cardinals, and no infinite successor ordinal is a cardinal (why?).

  • Finally this leaves 3 and 4. Suppose $\kappa=\aleph_\beta$ for some $\beta<\kappa$, for $\beta$ a limit; do you see how to write $\kappa$ as the sum of fewer-than-$\kappa$-many cardinals of size $<\kappa$? What about if $\beta$ is a successor? Now, why does this let you conclude that $\kappa=\aleph_\kappa$?

  • 1
    $\begingroup$ Thank you so much for the clarification! Yes, I was referring to strongly inaccessible but to have the definitions put succinctly as you have done really helps. Thank you also for the questions. To answer your question about successors, if kappa were a successor cardinal then I believe the power set of its immediate predecessor must have cardinality at least kappa, which contradicts inaccessiblity. As to limit ordinals, I see that my statement was redundant. EVERY infinite cardinal corresponds to a limit ordinal (actually an initial ordinal). I am working on your last question now. $\endgroup$ Commented Jan 9, 2017 at 14:56
  • $\begingroup$ I believe I understand your last point now. If Kappa were not a fixed point on the Aleph function then there would exist an alpha strictly less than kappa with kappa equal to Aleph alpha. But then the cofinality of kappa would be the cofinality of alpha (because kappa is a limit cardinal) which is less than kappa. This contradicts the regularity of kappa... I think I got that right? $\endgroup$ Commented Jan 9, 2017 at 21:29
  • $\begingroup$ @Sean Yup, that's right $\endgroup$ Commented Jan 9, 2017 at 21:44
  • $\begingroup$ Hi. Don't we also want to specify explicitly that $\kappa$ must be uncountable? (since $\aleph_0$ is a limit cardinal and is regular) $\endgroup$
    – PatrickR
    Commented Feb 28, 2022 at 8:05

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