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Given a set $\mathcal{C}$ of Cardinals, does $$D:=\bigcap_{C\in\mathcal{C}} C \in \mathcal{C}$$ hold? If $\mathcal{C}$ is finite or contains a finite cardinal, the answer is positive even for ordinals instead of cardinals because of elementary ordinal properties. The infinite case however does not seem so obvious.

Assuming the contrary, one could construct an infinite sequence of proper descreasing cardinals $C_0 \supsetneq C_1 \supsetneq C_2 \supsetneq \ldots \supsetneq D$ with all $C_k\in\mathcal{C}$. Intuitively I don't believe that's possible, not even for ordinals. I tried applying Zorn's lemma using $\supseteq$ for $\leq$ to give $D$ a proper structure, but using D as an upper limit with respect to this relation doesn't seem to work because the lemma seems to require $D\in\mathcal{C}$ to work, which is what I want to show. So that's a dead end for me.

I'm hardly used to ordinals and cardinals and grateful for help.

Background: I'm trying to do some formalization of (Multi)Graph theory in Mizar and given graphs with arbitrary vertex degree, I was wondering if it makes sense do differ between the concept of the infinum of all vertex degrees (i.e. their intersection) and the minimum (if present, smallest vertex degree).

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    $\begingroup$ Are you assuming the common representation of "cardinal" as "initial ordinal"? If not, then it doesn't necessarily make sense to speak of the intersection of a set of them. But if you do, you when you have is just a particular case of a set of ordinals, and such a set always has a least element (because the ordinals are well-ordered). $\endgroup$
    – hmakholm left over Monica
    Dec 2, 2017 at 1:59
  • $\begingroup$ "If"-case: thanks, I just thought of that possibility which led me to my own answer below. Good to know that this property holds for any well-ordered set. "If not"-case: I don't really know what you mean, but it seems it doesn't apply here. $\endgroup$
    – SK19
    Dec 2, 2017 at 2:18
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    $\begingroup$ For "if not", I was thinking about for example using Scott's Trick instead of ordinals to represent cardinals. In that case the sets that represent two different cardinals are always disjoint. Or, in general, we could take an axiomatic approach, and simply say that "cardinality of" is some class function $f$ on all sets such that $f(A)=f(B)$ exactly when $A$ and $B$ are equinumerous, and a "cardinal" is something that is in the range of that function. In this latter case, asking about the intersection of cardinals is not well defined. $\endgroup$
    – hmakholm left over Monica
    Dec 2, 2017 at 3:08

1 Answer 1

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Right after posting an idea came to mind. I didn't find anything about this topic on stack exchange or anything intersection related in the ORDINAL or CARD articles of the Mizar Mathematical Library, but I remembered the concept of minimum for sets of natural numbers and voila, found out that an equivalent holds for ordinals. (As Henning Makholm commented, this is because Cardinals are well-ordered.)

Therefore as long as the set $\mathcal{C}$ is not empty, it contains a minimum ordinal $E$, which therefore must equal the intersection of all cardinals for elementary ordinal properties: $D\subseteq E$ because $E\in\mathcal{C}$. On the other hand, for any $e\in E$ and $C\in\mathcal{C}$ holds $e\in C$ because $E\subseteq C$, hence $e\in D$ and $E\subseteq D$. Since $\mathcal{C}$ was required to contain only cardinals, $E=D$ must be a cardinal too.

So the answer is yes. Sometimes it really helps to think out loud :)

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