I was reading about ordered fields $\mathbb{F}$ with countable cofinality which means that there is a countable set $S$ that is cofinal. The definition of $S$ being cofinal is that for all $a \in \mathbb{F}$ there exists $s \in S$ such that $s \geq a$.

I was wondering about the meaning of the name "cofinal". It seems (at least to me) that it would imply that there is such a notion as a "final" subset. Oddly enough, the dual notion seems to be that of a coinitial subset (just switch the direction of the inequality in cofinality).

I was just wondering if anyone knew if there was any meaning to the "co-". I looked around on Wikipedia and did not really find anything. The closed that I came was the definition of an initial/lower set. So, I suppose that to dualize the definition of initial set, you take its definition: $\forall s \in S, (y \leq s \Rightarrow y \in S)$ and somehow turn it into coinitial: $\forall y\in \mathbb{F}, (\exists s \in S, s \geq y)$.

I do not exactly see how one systematically creates the notion of co- for coinitial sets. And especially so for cofinal sets.

Thank you for your help.

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    $\begingroup$ In this case, I think "co-" isn't used in the sense of duality, but rather as "together". A cofinal set is final with the base set, as in they reach equally far, "together". $\endgroup$
    – Arthur
    Oct 1, 2017 at 18:16
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    $\begingroup$ It could just as well be called "final" rather than "cofinal". This practice was adopted by category theorists when they generalised the notion of a cofinal set to that of a final functor. In part, I suspect this is because of the confusion with the "co-" prefix. (Indeed, the dual of a final functor is an initial functor, which is called 'co-cofinal' by some authors.) I suspect that set theorists / order theorists chose "cofinal" over "final", since "final subset" could be confused with "terminal segment" (i.e. upwards-closed subset). $\endgroup$ Oct 1, 2017 at 18:41
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    $\begingroup$ @Arthur: They're not the same in general. Consider the poset $\mathbb{R} \cup \{ \star \}$, where $\mathbb{R}$ has its usual order and $\star$ is not comparable to any $x \in \mathbb{R}$ (so $\star \le \star$ but, for all $x \in \mathbb{R}$, $\star \not\le x$ and $x \not\le \star$). In this poset, $\mathbb{R}$ is unbounded but not cofinal, since any cofinal subset must contain $\star$. $\endgroup$ Oct 1, 2017 at 19:12
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    $\begingroup$ @Arthur: They're also different in any totally ordered set with a maximal element. In fact, in such posets, no subset is unbounded, but all subsets containing the maximal element are cofinal. However, they do coincide in totally ordered sets with no maximal element. $\endgroup$ Oct 1, 2017 at 19:18
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    $\begingroup$ "co-" here means the same thing as in "co-founder" or "co-conspirator" or "co-author" or "co-owner". $\qquad$ $\endgroup$ Mar 4, 2018 at 0:28


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