# The supremum of any set of cardinals (considered as a set of ordinals) is again a cardinal.

An ordinal $\alpha$ is a cardinal iff no $\xi < \alpha$ is equivalent to $\alpha$. Now, let $A$ be any set of cardinals and $\sup(A)=\alpha$, then for $\xi < \alpha$ there is a $\beta \in A$ such that $\xi < \beta$. As such $|\xi|$ is smaller than that of $\beta$, and thus it is smaller than that of $\alpha$.

Does this sound correct/in the right direction? If so, could someone help better explain the first line?

• That's correct, I think. Not much more I can say about it, really - it's almost exactly what I would have written. – Patrick Stevens Feb 18 '16 at 12:52
• Note that you didn't handle the case where the supremum is just the maximum. It's the trivial case, but I thought it deserved to be mentioned. – Stefan Mesken Mar 1 '16 at 12:12

Fact 1: An ordinal $\alpha$ is a cardinal iff $\xi < \alpha \Rightarrow |\xi|<|\alpha|$. (This is the definition of cardinals; see, e.g., http://euclid.colorado.edu/~monkd/m6730/gradsets06.pdf)

Here $|\xi|$ denotes the cardinality of $\xi$.

Fact 2: The supremum of a set of ordinals is a well-defined ordinal, namely $\sup A:=\cup A$. See, e.g., Is the supremum of an ordinal the next ordinal?

Theorem the supremum of any set of cardinals is a cardinal.

Proof: Now, let $A$ be any set of cardinals and $\sup(A)=\alpha$. If $\max A$ exists, then $\alpha=\max A\in A$, hence then $\alpha$ is a cardinal.

If $\not\exists\max A$, then for any $\xi < \alpha$ there is a $\beta \in A$ such that $\xi < \beta$. But then $|\xi|<|\beta|\le|\alpha|$, by Fact 1.* As $\xi<\alpha$ was arbitrary, $\alpha$ is a cardinal, by Fact 1, QED.

++) In fact here $|\beta|<|\alpha|$, but we don't need that fact. (Proof: as otherwise for any $\gamma\in A$ we would have $|\gamma|\le|\alpha|=|\beta|$, hence $\gamma\le\beta$, as both are cardinals, so $\beta=\max A$, which was ruled out.)

(This is essentially the proof given above by others; I just wanted to make it clearer.)