TITLE: Does every ordinal have cardinality no greater than $\aleph_\mathbb{0}$?
QUESTION [6 upvotes]: My notes say that the ordinals $\omega + 1, \omega + 2, ... , 2 \omega, ... , 3 \omega, ... \omega^2, ... $ are all countable, and hence have cardinality equal to $\omega = \aleph_\mathbb{0}$. So I was wondering if it's fair to say that every ordinal has cardinality no greater than $\aleph_\mathbb{0}$?
Alternatively, I guess it's possible that that the sequence of infinite ordinals listed above, does not include some of the ordinals with a cardinality greater than the naturals.. but I wasn't sure.

REPLY [2 votes]: The ordinal $\omega_1$ has cardinality $\aleph_1$ which is greater than $\aleph_0$. Similarly, $\omega_2$, $\omega_3$,.... has cardinalities which are all greater than $\aleph_0$. 

You can say that every countable ordinal has cardinality which is not greater than $\aleph_0$, in fact the cardinality is equal to $\aleph_0$. 

Note that the ordinals $\omega + 1, \omega + 2, ... , 2 \omega, ... , 3 \omega, ... \omega^2, ... $ are all countable ordinal, however $\omega_1$ is the first uncountable ordinal.