0
$\begingroup$

My understanding is that any cardinality is always an integer because it expresses how many elements are in a given set. And I read that $\aleph_1$ is the next smallest cardinality that's larger than $\aleph_0$, so it seems to me that $\aleph_1 = \aleph_0+1$. However I don't see this equation anywhere on the Internet so I guess I'm wrong. Where am I wrong?

$\endgroup$
  • 2
    $\begingroup$ Only the finite cardinalities are (nonnegative) integers; the rest are, well, cardinals, which are different. $\aleph_0 + 1$ is just $\aleph_0$ again. $\endgroup$ – Qiaochu Yuan Jul 23 '18 at 7:18
  • $\begingroup$ What is true is that $\aleph_1=\aleph_{0+1}$. $\endgroup$ – Lord Shark the Unknown Jul 23 '18 at 7:19
  • $\begingroup$ $\aleph_{0+1}=\aleph_1\gt\aleph_0=\aleph_0+1.$ $\endgroup$ – bof Jul 23 '18 at 7:19
  • $\begingroup$ @QiaochuYuan Is there a special symbol or something to express the value of $\aleph_1$ - $\aleph_0$? $\endgroup$ – stacko Jul 23 '18 at 7:21
  • 1
    $\begingroup$ @stacko this is independent of $ZFC$(see en.wikipedia.org/wiki/Continuum_hypothesis) $\endgroup$ – Holo Jul 23 '18 at 7:34
0
$\begingroup$

The wrong part is that you took what you know about finite cardinals and extend it to infinite ones.

Assuming choice $\aleph_0$ is the smallest cardinal that is greater than all the finite cardinal, the first limit ordinal. To advance to $\aleph_1$ we need to find a set such that there is no injective from $A$ to $\Bbb N$, but $\aleph_0+1=|\Bbb N\cup\{0\}|=|\Bbb N|=\aleph_0$.

Even more: for $\kappa,\nu$ cardinals such that one of them is infinite we have $\kappa+\nu=\kappa\cdot\nu=\max\{\kappa,\nu\}$

$\endgroup$
  • $\begingroup$ @Arthur yes, but it is not the smallest without it $\endgroup$ – Holo Jul 23 '18 at 7:30
  • 1
    $\begingroup$ We only need countable choice for $\aleph_0$ to be the smallest infinite ordinal (and I think it's minimal regardless), but fair enough. We do need choice for $\max$ to be well-defined, though. $\endgroup$ – Arthur Jul 23 '18 at 7:36
  • $\begingroup$ @Arthur indeed even without choice every infinite cardinal is either not comparable or greater equal to $\aleph_0$ $\endgroup$ – Holo Jul 23 '18 at 7:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.