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Let $f : \mathbb{R}^m \rightarrow \mathbb{R}^n$ be a $C^1$ function. Prove or disprove that $\{x \in \mathbb{R}^m : f(x) = 0 \}$ is a closed set. How would you prove this??

I do not even understant what $f(x)=0$ represnets. I assume that it represents a surface in $\mathbb{R}^n$ but I am not sure.

eChung00
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2 Answers2

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I'll prove a more general result using the continuity of $f$. Call your set $A$. Take $x \in A^c$, i.e. $f(x)\neq0$ (the zero vector of $\mathbb{R}^n$).

By continuity there exists $\delta >0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<|f(x)|/2$ which gives by the triangle inequality $|f(y)|>|f(x)|/2$ for all $y$ in the open ball $B_\delta (x)$. Hence $B_\delta (x) \subset A^c$, which means that $A^c$ is open, and therefore $A$ is closed.

Reveillark
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  • I have a question. If you are given a question like this, how would you know if the set is closed? I mean, the only infomation give here is that $f$ is vector-valued differentiable function and the set. But I am confused about the decription of the set. I mean $f(x)=0$ what is this?? If I am give a set like ${(x,y) \in \mathbb{R}^2 : x^2 + y^2 < 2}$, then I can work with it. But if I am given a set described by a function with no formula, how would I determine if the set is closed?? – eChung00 Mar 28 '15 at 18:47
  • You are given an arbitrary function, you do not know the exact "shape" of the set, only that it is the set of all $x \in \mathbb{R}^m$ such that $f(x)=0$, i.e. all those $x$ that are mapped by $f$ to the zero vector. I gave proof above that the set $\left { x \in \mathbb{R}^m : f(x)=0 \right }$ is closed proving that its complement (those $x$ which are not mapped to $0$) is an open subset of $\mathbb{R}^m$ using only the continuity of the function $f$. Does this make sense? – Reveillark Mar 28 '15 at 19:17
  • Oh, ok. I was thinking that $f(x) = 0$ was an implicit function because my professor recently talked about an implicit function theorem and I remember she expressed a function in this way. So in this case, for all input vectors $x$ is mapped to zero vector. So since the set only contains a zero vector, we would know that the set is closed at first glance?? The question asks to prove or disprove. So it is hard for me to just start thinking about how to prove without the intuition that it might be closed. – eChung00 Mar 28 '15 at 19:51
  • Also I lost from the triangle inequality part of your proof. How did you get $|f(y)| > \frac{|f(x)|}{2}.$ And how did you conclude that every ball centered at $x$ with radius $\delta$ is contained in $A^c$?? This would be really basic question but I am really having hard time with this analysis class. – eChung00 Mar 28 '15 at 19:54
  • The notation is similar to that used in the implicit function theorem, but this particular problem has nothing to do with that. You'll get to that theorem in due time. For the triangle inequality I did: $|f(x)|=|(f(x)-f(y))+(f(y)|≤|f(x)-f(y)|+|f(y)|$ which gives $|f(x)|-|f(y)|≤|f(x)-f(y)|$. Since $|f(x)-f(y)|<{|f(x)| \over 2}$ it follows that $|f(y)|≥|f(x)|-|f(x)-f(y)|>{|f(x)| \over 2}$. You knew that $f(x)\neq 0$, hence $|f(x)|>0$, but then $|f(y)|>{|f(x)| \over 2}>0$. Hence $f(y)\neq 0$ for all $y \in B_{\delta}(x)$. But this means $B_{\delta}(x) \subset A^c$, by definition of $A$. – Reveillark Mar 28 '15 at 20:27
  • Cristal clear..!! Like * 10000000. Thank you very much! – eChung00 Mar 28 '15 at 20:55
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Well, $\{x\in\Bbb R^m: f(x)=0\}=f^{-1}(0)$. Now, because $f$ is continuous and $\{0\}$ is closed, $f^{-1}(0)$ is closed.

Tim Raczkowski
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