$f: \mathbb R^{n}\to \mathbb R^{n}$ ($n\geq 2$) is a continuous function.
Let $$M=\{x\in \mathbb R^n~|~f(x)=0\}$$
Is $M$ a closed set? If not, can you give a counter example.
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2Yes, assuming that singleton points are closed in the topology on $\mathbb{R}^n$. (They are in the standard metric topology.) – Sammy Black Mar 22 '13 at 06:39
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how about: $f^i:R^n\to R^n$ is a continuous function for $i=1,2$.Let $$M={(x,y)\in R^{2n} | f^1(x)+f^2(y)=0}$$ Is M a closed set? – Eric Brown Mar 22 '13 at 06:48
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1You ought to ask this as a separate question. The short answer is yes. – Sammy Black Mar 22 '13 at 08:42
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We have $M=f^{-1}({\{0\}})$ so $M$ is closed set of $\mathbb{R}^n$ as preimage of the closed set $\{0\}$ by the continuous function $f$.
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Yes. For all $y\neq 0$, there is a neighborhood $V$ of $y$ such that $0\notin V$. Then the preimage of this $V$ is necessarily an open set (by continuity), and it follows that $M^c$ is open, and therefore that $M$ is closed.
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