Recently I've came across such problem: give a polynomial $P(x,y)$, with $\inf_{\mathbb{R}^2} P=0$, but there is no point on the plane where $P=0$. I couldn't solve it after a day, and seriously doubt whether such a function exists, however its source claims that there is. Is that really possible?
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Is this in the real plane $\mathbb{R}^2$, or could it perhaps be over the rationals? – hardmath Jan 15 '13 at 19:51

The real plane. – aplavin Jan 15 '13 at 19:51

I edited "$in f(P)$ to $\inf f(P)$, as that seems to be the more common notation. Nice question, by the way. – Fredrik Meyer Jan 15 '13 at 19:51
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$P(x,y)=(1xy)^2+x^2$ has this property. Clearly $P>0$ and also the sequence $(x_n,y_n)=(1/n,n)$ shows that $\inf P=0$.
JP McCarthy
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3I read about it somewhere before  I'm not sure where exactly. If I recall correctly it is a counterexample to the claim that $\text{im } P(x,y)$ is closed. – JP McCarthy Jan 15 '13 at 20:05


Sorry. The image of the real plane under a polynomial in two variables. I suppose one would think it would be closed. – JP McCarthy Jan 15 '13 at 20:16