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We have $M$ an o-minimal structure. $X \in M^n$ with the induced topology. I'm reading an article which shows that $X \in M^n$ is definable compact is equivalent to $X$ being bounded and closed.

Definable compactness of $X$ means that any $M$-definable curve in $X$ is completable. (a curve in $X$ is a $M$-definable continuous embedding $f: (a,b) \rightarrow X$). It is said to be completable if $\lim_{x\rightarrow a^{+}}f(x)$ and $\lim_{x\rightarrow b^{-}}f(x)$ exists.)

When it shows that any definably compact subset $X \in M^n$ is bounded I've got the feeling that its proof is very complicated. I might be doing something wrong but I've got the feeling that this proof can be done much more easily.

Here is how it goes: it first shows that definable compactness is preserved under projection on the k first coordinates.

Then it proceeds by induction. let's assume that any definably compact subset $X \in M^n$ is bounded. Let $X \in M^{n+1}$, then as $p(X)$ is definably compact (by preservation of definable compactness under projection), it is by induction bounded ($p$ is the projection onto the $n$ first coordinates). So the first $n$ coordinates of X are bounded.

Now why can't we also say the projection of X onto his last n coordinates is definably compact (by preservation of definable compactness under projection), it is by induction bounded. So the last n coordinates of X are bounded.

So X is bounded.

(for the proof of "Let $S \in M^n$ be definably compact and let $p : M^n \rightarrow M^k$ be a projection map. Then p(S) is definably compact." is the following: By induction, it suffices to show that if $S \in M^{n+1}$ is definably compact then p(S), where $p: M^{n+1} \rightarrow M^n$ denotes projection onto the first n coordinates, is definably compact as well. For a contradiction, assume not. Then there is a definable continuous embedding $f: (a, b) \rightarrow p(S)$ such that,say, f does not have a right-hand limit point in $p(S)$. By o-minimality, for every $a \in p(S)$, the set $S_a = \{b \in M : (a, b) \in S\}$ is the union of finitely many intervals. Since S is definably compact, $S_a$ is closed and bounded. Let m(a) be the least element of $S_a$. We now define $g: (a, b) \rightarrow S$ by $g(x) = (f(x),m(f(x))$. It follows that g does not have a right-hand limit point in S since f does not have one in $p(S)$, a contradiction.)

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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. –  David Roberts May 31 '11 at 12:14
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This argument sounds OK to me. In fact, I think you don’t even need induction: just project $X$ on each coordinate individually. –  Emil Jeřábek May 31 '11 at 12:15
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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. –  Thierry Zell May 31 '11 at 13:03
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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 –  ACL May 31 '11 at 13:25
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@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)'. –  ACL May 31 '11 at 14:51
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