If $\kappa_i, i\in I,$ are distinct cardinal numbers, then $|I|\leq\sup\{\kappa_i\mid i\in I\}$

Question

It's used in the book I'm reading without even being mentioned explicitly. It seems intuitive, but I cannot prove it.

Answer 1

Presumably $I$ or at least some $\kappa_i$ are infinite.

Let $\lambda=\sup\{\kappa_i\mid i\in I\}$, then $\lambda$ is an infinite ordinal. So $I\mapsto\lambda+1$, the ordinal, by mapping $i\mapsto\kappa_i$. Since for an infinite ordinal, $|\lambda+1|=|\lambda|$, we are done.

If you want to be a bit more specific about it, you can note that since the $\kappa_i$ are cardinals, if the above happens, then there is some "gaps" between them (i.e. not all the cardinals are finite, or $\{\kappa_i\mid i\in I\}=\omega+1$ in which case we can manually move everything around) so we can put the maximum cardinal, if it exists, into one of these gaps.

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Finally, note that if $I=\{0\}$, and $\kappa_i=0$, then $\sup\{\kappa_i\mid i\in I\}=0$, but $|I|=1$. So some assumption of infinitude is necessary here.