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Let $f$ be a surjective continuous function from $\Bbb{R}^n$ to $\Bbb{R},$ I look at the set $$A:=\{x\in \Bbb{R}^n:f(x)=0\}.$$

$A$ is closed, but I wonder if it can be compact, for $n=1$ it can. But what's happen if $n\ge 2$?

Is it possible to find such a function where the set of zero is bounded?

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    The zero set may be bounded $f(x,y)=x^2+y^2$ or not : $f(x,y)=x$. – 2017-01-08
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    So many hasty answers... why ? – 2017-01-08
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    You're seriously complaining about people trying to help you? – 2017-01-08
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    @MichaelHoppe: either one in your comment would not work: the first one is not onto and the second one has $f^{-1}\{0\}$ being unbounded. – 2017-01-08
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    @JorgeFernándezHidalgo Seriously yes, I said "hasty". How can you think that it implies that I am complaining about helping me ? – 2017-01-08
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    He seems to be complaining about people jumping to post an answer without properly reading the question. That is a fair criticism which I share. The reason is obvious: people feel a pressure to rush out an answer before getting "scooped". If you stop and properly think about it someone else might post before you. Thus questions like this which at first glance looks very simple but turns out to be a bit subtle tend to get alot of wrong answer. It's just a biproduct of how the point-system works. Anyway, it's no big problem imo, all the wrong answers have been deleted. – 2017-01-08

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For $n > 1$, the zero set of a continuous surjective $f \colon \mathbb{R}^n \to \mathbb{R}$ is always unbounded.

Consider an arbitrary $r \geqslant 0$. The set $f(K_r(0))$, where $K_r(x) = \{ y : \lVert y-x\rVert \leqslant r\}$, is compact, hence $f$ attains negative values and positive values on $U_r := \mathbb{R}^n\setminus K_r(0)$. Since $U_r$ is connected, $f^{-1}(0)\cap U_r \neq \varnothing$.

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    Nice, how did you get the idea? – 2017-01-08
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    the first idea is to try to make the set of zeros a closed curve and make the interior take on the negative values, but one easily sees that won't work. Analyzing why that happens leads to the solution, or at least that's how I think of it. – 2017-01-08
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    Pretty much what Jorge said. One takes a continuous function with compact zero set and thinks about the consequences. One consequence is that such a function must be bounded above or below, hence cannot be surjective. – 2017-01-08