Point of $\kappa$-completeness of an ultrafilter when defining measurable cardinals

Question

When defining the notion of a measurable cardinal using the ultrafilter definition, why do we require the ultrafilter to be $\kappa$-complete? I know this makes our definition of a measurable cardinal more restrictive (by requiring more sets to be in the ultrafilter) but intuitively speaking, what effect does this condition have on the "size" of kappa? What would happen to our definition if we left it out?

Answer 1

"When defining the notion of a measurable cardinal using the ultrafilter definition, why do we require the ultrafilter to be $\kappa$-complete? I know this makes our definition of a measurable cardinal more restrictive (by requiring more sets to be in the ultrafilter)..."

It's not correct to say that $\kappa$-completeness requires more sets to be the in the ultrafilter. It's true that $\kappa$-completeness means that the intersection of fewer than $\kappa$ many sets in the ultrafilter is also in the ultrafilter — but you could just as easily say that $\kappa$-completeness requires more sets not to be in the ultrafilter because the union of fewer than $\kappa$ many sets not in the ultrafilter is also not in the ultrafilter.

It's best to think of $\kappa$-completeness as simply a structural requirement on the ultrafilter as a whole; it doesn't mean that more sets or fewer sets are in the ultrafilter than in a non-$\kappa$-complete ultrafilter, just different sets.

"... but intuitively speaking, what effect does this condition have on the "size" of kappa? What would happen to our definition if we left it out?"

If the alternative is just to require $\kappa$ to be an uncountable cardinal that carries a non-principal ultrafilter (with no completeness requirement at all), then every uncountable cardinal satisfies the property. So this isn't a large cardinal property at all (or even a newly interesting property).

If you replace $\kappa$-completeness by countable completeness, then the cardinals satisfying this property are precisely those cardinals greater than or equal to the first measurable cardinal. So, for $\kappa$ greater than the first measurable cardinal, this doesn't imply that $\kappa$ is dramatically larger than the cardinals less than $\kappa.$

With the standard definition of measurability, every measurable cardinal is much larger than the cardinals less than it. That's why it has been adopted as the definition of measurability.

Answer 2

Your question leaves room for interpretations and may therefore not have a definitive answer. Thus, instead of trying to come up with one, let me highlight two factors such an answer would have to take into account.

First, there is a historical reason. The definition of a measurable cardinal is ultimately due to Stanislaw Ulam, who at the age of 21 proved in Zur Masstheorie der allgemeinen Mengenlehre (in German) the following fact.

Theorem [Ulam]. If there is a $\sigma$-additive nontrivial measure on $X$ then there is a a two-valued measure on $X$ and $\operatorname{card}(X)$ is greater than or equal to the least inaccessible or there is an atomless measure on $2^{\aleph_0}$ and $2^{\aleph_0}$ is greater than or equal to the least weakly inaccessible.

Combine this with the the following two facts:

Lemma. If $\kappa$ admits a two valued measure $\mu$, then there is a $\sigma$-complete ultrafilter $U_\mu$ given by $$ X \in U :\iff \mu(X) = 1 $$ and

Lemma. Let $\kappa$ be the least cardinal for which there is a $\sigma$-complete ultrafilter $U$ on $\kappa$. Then there is a $\kappa$-complete ultrafilter on $\kappa$.

and you see that the least candidate for a measurable cardinal will in fact admit a $\kappa$-closed ultrafilter.

Second, let $\lambda$ be an inaccessible cardinal and let $U$ be an ultrafilter on $\lambda$. Suppose that $U$ is at least $\sigma$-closed and let $\kappa$ be least such that $U$ is not $\kappa^+$-closed, i.e. there is a sequence $(X_\alpha \mid \alpha < \kappa) \in ^\kappa U$ such that $\bigcap_{\alpha < \kappa} X_\alpha \not \in U$. (Note that $\kappa \le \lambda$.) We may form an elementary embedding $$ j_U \colon V \to \operatorname{Ult}(V;U) $$ such that $\operatorname{Ult}(V;U)$ is transitive and one can show $\kappa$ is the least ordinal $\alpha$ with $j_U(\alpha) \neq \alpha$. We say that $\kappa$ is the critical point of $j_U$. This allows us to define a $\kappa$-closed ultrafilter $W$ on $\kappa$ by $$ X \in W \iff \kappa \in j_U(X). $$ In particular, we may now form $$ j_W \colon V \to \operatorname{Ult}(V; W). $$ But from $j_W$ we can recover a $\kappa$-closed ultrafilter $\bar{U}$ on $\lambda$ by letting $$ X \in \bar{U} \iff \kappa \in j_W(\bar{U}). $$

Thus, if $\kappa < \lambda$, then the fact that $\lambda$ admits a $\kappa$-closed ultrafilter really doesn't have anything to do with $\lambda$. Instead it is an consequence of the fact that $\kappa$ admits a $\kappa$-closed ultrafilter. A slight variations of this also shows that any $\mu > \kappa$ admits a $\kappa$-closed ultrafilter and therefore the reason that all of these $\mu > \kappa$ have a $\kappa$-closed ultrafilter on them is really due to the measurability of $\kappa$.

You can find a proof of all of the above facts and much more about measurable cardinals in Jech's wonderful textbook 'Set Theory (The 3rd millenium edition)'.