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Dec 25, 2011 at 16:54 history edited Goldstern
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Sep 6, 2011 at 21:46 history edited j.c. CC BY-SA 3.0
edit title
May 31, 2011 at 14:51 comment added ACL @Thierry: Why not? :-) Well, google pointed at it,this paper is the one where the definable compactness is introduced and shown to be equivalent to bounded & closed, and the proof is exactly as indicated by `unknown (google)'.
May 31, 2011 at 13:43 comment added Thierry Zell @ACL: I'm just curious: why this paper in particular?
May 31, 2011 at 13:34 comment added ACL Anyway, you need to prove that a subset of $M$ is definably compact if and only if it is bounded and closed. But this follows from the fact that definable functions $(a,b)\to M$ are piecewise monotonic.
May 31, 2011 at 13:25 comment added ACL I presume the paper is Definable Compactness and Definable Subgroups of o-minimal groups, by Ya'acv Peterzil and Charles Steinhorn. Journal of the London Mathematical Society (1999), 59: 769-786
May 31, 2011 at 13:03 comment added Thierry Zell It would help if you actually referenced which article you are reading; this proof appears in many places. Your suggested proof appears to work, but your question does not really tell us what the "complicated" proof in the original paper is.
May 31, 2011 at 12:54 history edited Thierry Zell CC BY-SA 3.0
rightarrow
May 31, 2011 at 12:15 comment added Emil Jeřábek This argument sounds OK to me. In fact, I think you don’t even need induction: just project $X$ on each coordinate individually.
May 31, 2011 at 12:15 comment added David Roberts I also added a tag.
May 31, 2011 at 12:14 history edited David Roberts
Added tag
May 31, 2011 at 12:14 comment added David Roberts This is a decent question, but can I suggest editing it and changing the title? With the title as is (and with no capital), it may not generate the interest it deserves.
May 31, 2011 at 11:57 history asked user15496 CC BY-SA 3.0