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We sometimes need to distinguish the counted from the countable. A counted set is a set equipped with a particular bijection into (some of) the natural numbers; a set is countable if there exists such a bijection. One way the difference shows up: to show that union of countable sets $X_i$ is countable you need some choice; not so to show that union of counted sets $X_i$ is countable. [I learnt this way of talking from Thomas Forster, who got it from John Conway. And I'm reminded of this by a very recent comment here to a question about Proving existence of a surjection $2^{\aleph_0} \to \aleph_1$ without AC, where Peter LeFanu Lumsdaine says "This is fairly standard terminology for working with countability without choice".]

OK, here's a further terminological question. There's a parallel distinction to be drawn between the case where we have (say) two ordered sets equipped with a particular order-preserving bijection between them, and the case where we are just given that the two ordered sets are order-isomorphic. Being order-isomorphic (like being countable) is defined in terms of a quantification over bijections. Is there a standard bit of terminology for the corresponding notion (like being counted) where the relevant sets are taken together with a particular bijection?

[Of course, there will a distinction like this to be drawn whatever kind of isomorphisms we happen care about.]

Added A little context. The notion of being countable, defined as it is in terms of a quantification over functions, might seem to be bit problematic if used in a context where we are being non-committal about what functions are being countenanced (and so non-committal about what the domain of quantification is). Similarly for the notion of being isomorphic, defined as that is in terms of another quantification over functions.

But in fact, often we don't have to fuss very much about such matters in the first case, because we are typically interested in the countable-because-actually-counted, i.e. we actually equip the relevant set with a bijection into the numbers. Likewise, we typically establish that two ordered sets are order-isomorphic by actually giving a bijection that is order-preserving: we are interested in the isomorphic-because-actually-.... Ah, what's word we want here?

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Peter Smith
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  • I don't personally know of such, but couldn't you drop the sets and just deal with the isomorphisms? – dfeuer Aug 23 '13 at 16:28
  • The same occurs in many places. For example $\mathbb R[X]/(X^2+1)$ is isomorphic to $\mathbb C$, but there is no canonical isomorphism ... – Hagen von Eitzen Aug 23 '13 at 16:36
  • @HagenvonEitzen I am unsure what you mean. Do you mean just because there are 2 isomorphisms ($x\rightarrow -i$ or $i$)? I think Peter wants isomorphisms that are not really constructable without choice, but can be proven to exist with choice. A better example for fields would be an isomorphism $\Bbb C(x)\rightarrow \bar \Bbb C$ which can be proven to exist but really cannot be explicitly described without some infinite choice. – PVAL-inactive Aug 23 '13 at 16:55
  • The isomorphism in my comment is supposed to be $\Bbb C\rightarrow\overline {\Bbb C(x)}$ – PVAL-inactive Aug 23 '13 at 17:02
  • Almost no one would consider using the word in a mathematical context like this, but you could use "isomorphed" – Mark S. Jan 01 '14 at 16:09
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    You could indeed ... Maybe I could learn to live with isomorphed! – Peter Smith Jan 01 '14 at 16:50

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