why Why is this proof so complicateddefinable compact equivalent to bounded and closed for sets with o-minimal structures?

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edited May 31, 2011 at 12:54

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We have M$M$ 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$X$ being bounded and closed.

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

When it shows that any definably compact subset X $\in M^n$$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 coordonatescoordinates.

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)$p(X)$ is definably compact (by preservation of definable compactness under projection), it is by induction boudedbounded (p$p$ is the projection onto the n$n$ first coordonatescoordinates). So the first n coordonates$n$ coordinates of X are bounded.

Now why can't we also say the projection of X onto his last n coordonatescoordinates is definably compact (by preservetionpreservation of definable compactness under projection), it is by induction boudedbounded. So the last n coordonatescoordinates 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 -> M^k$$p : M^n \rightarrow M^k$ be a projection map. Then p(S) is definably compact." is the following: By induction, it suffices to show that if $S \in M^{n+1}$ is definably compact then p(S), where p: $M^{n+1} -> M^n$$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) -> p(S)$f: (a, b) \rightarrow p(S)$ such that,say, f does not have a right-hand limit point in p(S)$p(S)$. By o-minimality, for every a $\in$(S)$a \in p(S)$, the set Sa = {b $\in$ M : (a, b) $\in$ S}$S_a = \{b \in M : (a, b) \in S\}$ is the union of finitely many intervals. Since S is definably compact, Sa$S_a$ is closed and bounded. Let m(a) be the least element of Sa$S_a$. We now define g: (a, b) -> S$g: (a, b) \rightarrow S$ by g(x) = (f(x),m(f(x))$g(x) = (f(x),m(f(x))$. It follows that g does not have a right-hand limit point in S since f does not have one in p(S)$p(S)$, a contradiction.)

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)-> X where (a,b) $\in$ M). 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 coordonates.

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 bouded (p is the projection onto the n first coordonates). So the first n coordonates of X are bounded.

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

So X is bounded.

(for the proof of "Let $S \in M^n$ be definably compact and let p : $M^n -> M^k$ 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: $M^{n+1} -> M^n$ 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) -> p(S) such that,say, f does not have a right-hand limit point in p(S). By o-minimality, for every a $\in$(S), the set Sa = {b $\in$ M : (a, b) $\in$ S} is the union of finitely many intervals. Since S is definably compact, Sa is closed and bounded. Let m(a) be the least element of Sa. We now define g: (a, b) -> 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.)

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.)