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There are several definitions of "finite", like Dedekind's and Tarski's (Thanks to A.K. for point out the latter - first time I've heard of it): From the Wikipedia entry on Finite Set:

(Richard Dedekind) Every one-to-one function from S into itself is onto.

(Alfred Tarski) Every non-empty family of subsets of S has a minimal element with respect to inclusion.

And there are more.

Do these distinctions matter when considering definitions of fundamental categories like topological spaces or measure spaces? For example,

In topology: Why do we require a topological space to be closed under finite intersection?

In measure theory: "A measure is continuous from above if [given measurable sets and closed under intersection] at least one set has finite measure," alternatively, in the definition of sigma-finite measures (same article).

Do the various definitions of finiteness lead to non-isomorphic or non-equivalent categories?

My question is not specifically about topology or measure theory, but these are basic definitions introduced at undergrad level, so I thought, better to understand the context via basic examples.

alancalvitti
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  • For those of us not in the know, can you provide a
  • reference to these definitions of "finite" per Dedekind or Tarski? 2) Not being familiar with either of these definitions, is there any example of a set which is "finite" in one definition and not in another?

    – NKS Jan 23 '13 at 04:54
  • @NKS, added Wikipedia links – alancalvitti Jan 23 '13 at 04:58
  • @NKS I believe there are (at least) 16 different definitions of 'finite set' and no two are equivalent. – Ittay Weiss Jan 23 '13 at 04:59
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    It should be pointed out that all these definitions are equivalent under the axiom of (countable) choice. – Asaf Karagila Jan 23 '13 at 07:47
  • @AsafKaragila, although I find Choice interesting, I am primarily interested in constructive mathematics. Green & Tao in their arithmetic progressions paper write that they carefully avoid Choice, like a plague of some sort. – alancalvitti Jan 23 '13 at 21:45
  • alan, I wrote this comment to point to the general public why these definitions are not the same. I might post an answer of my own later. I am still contemplating this. – Asaf Karagila Jan 23 '13 at 21:49
  • How many of the 16 definitions equivalent in ZF+AD? – alancalvitti Jan 23 '13 at 21:55
  • As far as I recall AD proves countable choice, so all Dedekind-finite sets are finite, and so all are equivalent. I should also point out that philosophically ZF+AD is a much harsher assumption than ZFC. I also don't really recall 16 definitions of finite. I recall about seven, but it's easy to generate more. – Asaf Karagila Jan 23 '13 at 23:50