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So I am having trouble doing the mapping in this problem. I fail to understand what is being mapped. Are both $\mathbf{x}$ and $f(\mathbf{x})$ in the $U$?

I see that I must show $n+1$ here, should I employ induction?

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The $n+1$ comes from the "graph" of $f$, where the first $n$ coordinates are for the $x$ in $R^n$ and the last coordinate is for the value $f(x)$ which is a real number. Only the $x$ is in $U$, since $U$ is said to be an open set in $R^n$. –  coffeemath Nov 14 '12 at 2:01
    
If you draw it for $n=1$ or $n=2$, that might help you understand the general situation. –  Mark McClure Nov 14 '12 at 2:07
    
OKay I'll heed all of your advices. Thank you –  jip Nov 14 '12 at 2:10
    
Is there any particular reason that you changed $f$ to $g$, $U$ to $K$? Those are just cosmetic changes--no big deal. The bigger quesion is, why did you remove the requirement that the function be real-valued? That's quite necessary if you want the claim to hold. –  Cameron Buie Nov 14 '12 at 2:22

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In this case, they're taking $\mathbf{x}=\langle x_1,...,x_n\rangle \in\Bbb R^n$, $f(\mathbf{x})\in\Bbb R$. When they say $\langle \mathbf{x},y\rangle $, that's a compressed way of writing the ordered $(n+1)$-tuple $\langle x_1,...,x_n,y\rangle \in\Bbb R^{n+1}$, not an ordered pair. The $n+1$ is coming from there--we won't be doing any induction.

Define $g:U\times\Bbb R\to\Bbb R$ by $g(\mathbf{x},y)=y-f(\mathbf{x})$. Then $g$ is a continuous function (can you see why?), and so the preimage of the open ray $(0,\infty)$ under $g$ is open in $U\times\Bbb R$. That is, $$g^{-1}\bigl((0,\infty)\bigr)=\bigl\{\langle\mathbf{x},y\rangle:\mathbf x\in U,y-f(\mathbf{x})>0\bigr\}=\bigl\{\langle \mathbf x,y\rangle :\mathbf x\in U,y>f(\mathbf x)\bigr\}$$ is open in $U\times\Bbb R$. Since $U$ is open in $\Bbb R^n$, then $U\times\Bbb R$ is open in $\Bbb R^n\times\Bbb R=\Bbb R^{n+1}$. Therefore, the desired set is open in $\Bbb R^{n+1}$.

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That last equality looks wrong to me. The left-hand side depends only on $U$ and $f$ but the right-hand side also depends on $y$. –  Math536 Nov 14 '12 at 2:43
    
@Math536: Crud! You're right. Thanks. I think I've fixed it, now. –  Cameron Buie Nov 14 '12 at 2:44
    
You mind if I just ask you why you chose $(-\infty, y)$? –  jip Nov 14 '12 at 6:53
    
@sizz: Because saying $f(\mathbf{x})\in(-\infty,y)$ is equivalent to saying $y>f(\mathbf{x})$. –  Cameron Buie Nov 14 '12 at 7:03
    
Could you be more explicit in explaining why the two are equivalent? I am not following at all –  jip Nov 14 '12 at 7:11

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