8
$\begingroup$

Consider a compact (semialgebraic) ball $B\subset \mathbb{R}^n$ and a semialgebraic set $A=A(f_1,\cdots,f_s)\subset \mathbb{R}^n$ defined through some representation in terms of polynomials $f_1,\ldots,f_s\in \mathbb{R}_m[x_1,\ldots, x_n]$.

Is the dependence of $A(f_1,\cdots,f_s)\cap B$ on the polynomials $f_1,\cdots,f_s$ outer semicontinuous?

Are there known results like this in the theory of real semialgebraic sets?

A set-valued function $F:X\to \mathcal{P}(\mathbb{R}^n)$, that to each $x\in X$ associates a compact subset $F(x)\subset\mathbb{R}^n$, is called outer semicontinuous when for every $x_0\in X$ and $\delta>0$ there exists a neighborhood $U$ of $x_0$ such that $F(x)\subset N_\delta(F(x_0))$ for all $x\in U$.

$N_\delta(A)$ stands for a $\delta$-neighborhood of a set $A\subset \mathbb{R}^n$.

$\endgroup$
  • 3
    $\begingroup$ Can you define the concept of outer semicountinuity of a familly of sets? $\endgroup$ – Liviu Nicolaescu Sep 26 '15 at 22:58
  • $\begingroup$ Which topology do you consider on the space of polynomials? The number $s$ is probably fixed. What about the degrees? $\endgroup$ – Jochen Wengenroth Sep 30 '15 at 13:29
  • $\begingroup$ Sorry for the slopiness, I meant s fixed and the degrees of the polynomials bounded, so that the topology is clear. $\endgroup$ – p. duarte Oct 1 '15 at 10:54
1
$\begingroup$

This seems like it can't be true since the sets $A(f_1, \ldots, f_s)$ may not be compact, which violates the precondition for outer semicontinuous, but also makes the postcondition not hold.

E.g., consider $C = \{(x, y): xy-1 \ge 0\}$. It does not seem like we can force $A(f) \subseteq N_\delta(C)$ for all $f$ near $xy-1$. You will have to use some $f \in U$ containing an $x$ term, which will already make $A(f)$ very different from $A(xy-1)$.

$\endgroup$
  • $\begingroup$ Sorry, the compactness assumption was implicit in my question. I have reformulated it to make it explicit. $\endgroup$ – p. duarte Sep 30 '15 at 9:24

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.