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

When it shows that any definably compact subset X $\in X \in M^n$ 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 bouded bounded (p $p$ is the projection onto the n $n$ first coordonates)coordinates). 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 coordonates coordinates is definably compact (by preservetion preservation of definable compactness under projection), it is by induction boudedbounded. So the last n coordonates 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: $M^{n+1} -> p: M^{n+1} \rightarrow M^n$ 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: $f: (a, b) -> p(S) \rightarrow p(S)$ 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 $\in$ S_a = \{b \in M : (a, b) $\in$ S} \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: $g: (a, b) -> S \rightarrow S$ by g(x) $g(x) = (f(x),m(f(x)). f(x),m(f(x))$. 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.)