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I have result: measure of the set of critical values of $f$ is zero (by Sard's theorem), where $f: \mathbb{R^n} \rightarrow \mathbb{R}$ are polynomial functions. How do you show that the set of critical values of $f$ is finite?

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    Wow, well done! Persistence is the key to success!2012-12-30

1 Answers 1

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$V=crit(f)_\Bbb C=(\cap_{i=1}^n\{x\in \Bbb C^n|\frac {df}{dx_i}(x)=0\})$

is an algebraic variety. Hence it is a finite union of irreducible algebraic varieties . If a variety is irreducible, it is easy to see it is connected in the Zariski topology. So in particular, $V$ has a finite number of connected components in the Zariski topology. An algebraic variety (in an algebraically closed field) is connected in the Zariski topology iff it is connected in the analytic topology (see here). Since $Df=0$ on $V$ and $f$ is smooth, we have $f$ is constant on each connected (in the analytic sense) component of $V$. Therefore $f$ only takes finitely many values on $V$. But $crit(f)_\Bbb R\subset crit(f)_\Bbb C$, so $f$ only takes finitely many values when restricted to its real critical set. So $f(crit(f)_\Bbb R)$ is finite as desired.