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Given an algebraic map $f: B^d \to \mathbb{R}$, from the unit ball of dimension $d$ to the real, let $Y = f^{-1}(0)$. Then it is always possible to find a smaller ball $B_r \subset B^d$ not necessarily centered at $0$, such that for all lines $l$ going through $B_r$, $|l \cap Y| \le k$, where $k$ is the maximum degree of $f$.

Is the analogue of this for higher codimension still true? By that I mean if $f: B^d \to \mathbb{R}^k$, $k \le d$, and q is a regular point of $f$. Then is it true that we can always find $B_r$ such that for all affine $n$-planes going through $B_r$, the number of intersections of the plane and $Y = f^{-1}(q)$ is bounded by the maximum degree of $f$, or perhaps some other polynomial bound depending only on the max degree of $f$?

I feel this should be doable using basic calculus of one variables, eg. mean value theorem. But it seems complicated.

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up vote 5 down vote accepted

Unless I misunderstood the question, here is a counter-example: $d=3$, $k=2$, $f:\mathbb R^3\to\mathbb R^2$ is just a linear map, e.g. $f(x,y,z)=(x,y)$. Take $q=(0,0)$, then $Y=f^{-1}(q)$ is a straight line in $\mathbb R^3$.

A generic 2-plane intersects $Y$ at one point. However there are planes containing $Y$ (and hence having infinitely many intersections), moreover these planes cover the entire space. So there is no ball $B_r$ avoided by such planes.

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You are so right. I need to reformulate the condition about the ball $B_r$. – John Jiang Nov 9 '10 at 7:05

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