polynomial function is a proper map
where is the degree of .
Recall that is compact if and only if is closed and bounded. Since polynomial functions are continous, it is sufficient to show that preimage of a bounded set is bounded. So assume that is bounded and is not bounded. Take a sequence such that
where denotes the Euclidean norm of .
Recall that for any we have . Thus we have:
Then is a real polynomial of degree and the leading coefficient of is positive, which implies that
Now for each we have
but tends to infinity, therefore tends to infinty. Contradiction, since for each we have that and is bounded.
Corollary 1. Polynomial functions on are closed maps.
Proof. Note that is compactly generated Hausdorff space and therefore every proper and continous map is closed. Thus (due to proposition) polynomial functions are closed.
Corollary 2. Assume that is a polynomial function such that for any . Let be a map defined by the formula
Then is bounded.
Proof. We wish to show that there exists such that for all the inequality holds. Since polynomial functions are closed maps, then the image of is a closed subset of . Therefore is open and it contains , thus there exists such that the ball around with radius has empty intersection with . This means that for all we have that . Now for and for any we have:
which completes the proof.
|Title||polynomial function is a proper map|
|Date of creation||2013-03-22 18:30:49|
|Last modified on||2013-03-22 18:30:49|
|Last modified by||joking (16130)|