If $f$ is a continuous function from $\mathbb R^2 \rightarrow \mathbb R$ such that $f(x)=0$ for only finitely many values of $x\in\mathbb R^2$. Can we conclude that $f(x)\leq 0$ or $f(x)\geq 0$ for all $x$?
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1Is $f$ a function $\mathbb{R} \to \Bbb R$ or ${\Bbb R}^2 \to \Bbb R$? If the latter, why does $f$ only take one argument? – Bruce Zheng Oct 16 '14 at 04:29
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2@BruceZheng I think $x$ is meant as a vector. – Oct 16 '14 at 05:06
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Yes, since otherwise there would be two points $x_1, x_2 \in \mathbb{R}^2$ with $f(x_1) < 0 < f(x_2)$. These can be joined by a smooth (or piecewise linear) path $\gamma$ which does not pass through the zeros. This contradicts the intermediate value theorem for $f$ restricted to $\gamma$.
Lukas Geyer
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The set $f^{-1} \{0\}$ is finite and so $\mathbb{R}^2 \setminus f^{-1} \{0\}$ is connected. Hence $f(\mathbb{R}^2 \setminus f^{-1} \{0\} ) $ is connected, and so must lie completely in either $(-\infty,0)$ or $(0,\infty)$.
Note: This is true if $f^{-1} \{0\}$ is countable., see Arcwise connected part of $\mathbb R^2$.
copper.hat
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