# For a given set $S$ , is it possible to find a well ordered set $T$ such that every $S$ indexed subset of $T$ has a strict upper bound?

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.

• My edit: 1st line : weather = whether. – DanielWainfleet Apr 23 at 3:20

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.
• 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 $T=(\omega_1,\in)$. – DanielWainfleet Apr 23 at 3:16