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I have recently been thinking about whether it is possible to construct a free algebra on a signature containing an operation of nonfinite arity. This lead me to propose a question to myself:


  1. For any set S, is it possible to find a well ordered set $ \mathbb T:= (T, <)$, such that for all functions $g: S \to T$ their exists an element $\alpha \in T$ such that $\alpha$ dominates $g[S]$ (i.e. $\forall x \in g[S]: \alpha > x$)?

It is a quick remark to note that any such $T$ must satisfy $|T| >|S|$ otherwise we have a surjective function $ f: S\twoheadrightarrow T $ of which we can find no such dominating $\alpha_f$ for!

I have been attempting to no avail to prove that such $\mathbb T$ exists for any set $S$ for a while now to no avail, and I am not certain that one necessarily does.

I have spotted an analogy between question (1) and having to take the existence of an infinite carnality set as an axiom in set theory. Note that the existence of $\omega$ is equivalent to asking if such a $\mathbb T$ exists when $S$ is a singleton set. This makes me wonder weather the totally ordered set I dream of is too big to exist if $S$ is of infinite cardinality, without making more assumptions than one usually does ($ZFC$), when doing set theory.


Therefore, I open the question (1) up to the floor.

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  • $\begingroup$ My edit: 1st line : weather = whether. $\endgroup$ – DanielWainfleet Apr 23 at 3:20
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Yes. The notion you're looking for is called cofinality. The cofinality of a well-ordered set $T$ is the smallest order type of an unbounded set.

So for example, if $T$ has a maximal element $t$, the smallest unbounded set is $\{t\}$, and so the smallest order type is $1$. But for $T=\Bbb N$ (with the usual order), since every finite subset is bounded, and every proper initial segment is finite, the only order type of unbounded sets is $\omega$, the order type of $\Bbb N$.


We say that a well-ordered set $T$ is regular if every unbounded subset is isomorphic to $T$. We can prove that cofinality is always regular. Moreover, if $\alpha$ is an ordinal which is regular, then it is a cardinal, and there is a proper class of well-orders whose cofinality is $\alpha$.

So the question is, is there a proper class of regular ordinals? Well, the answer is yes, assuming $\sf ZFC$. But it is consistent without choice that every well-ordered set has an unbounded set of order type $\omega$, which is truly an odd mathematical universe.

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  • $\begingroup$ To the proposer: If $k$ is an infinite cardinal ordinal then $k^+$ is the least cardinal ordinal greater than $k.$ And $k^+$ is regular. For example, $\omega^+=\omega_1,$ so if $S$ is countable let $T=(\omega_1,\in)$. $\endgroup$ – DanielWainfleet Apr 23 at 3:16

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