Is $\aleph_1= \aleph_0+1$ wrong?

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?

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

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 $\kappa+\nu=\kappa\cdot\nu=\max\{\kappa,\nu\}$
• 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. – Arthur Jul 23 '18 at 7:36
• @Arthur indeed even without choice every infinite cardinal is either not comparable or greater equal to $\aleph_0$ – Holo Jul 23 '18 at 7:59