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This is from Apostol's Calculus, Vol. II Section 8.3 #3(a).

Prove that $S=\{(x,y,z) \mid z^2 - x^2 - y^2 -1>0\}$ is open.

The only way to prove that a set is open which has been covered so far is to prove that for an arbitrary point $\mathop a\limits^{\small \to}   \in S$ there is an open ball $B(\mathop a\limits^{\small \to}  , r) \subset S$ by finding an explicit $r$ and showing that $\mathop x\limits^{\small \to}   \in B(\mathop a\limits^{\small \to}  , r) \implies \mathop x\limits^{\small \to}   \in S$. Nothing topological (compactness, completeness, etc.) has been discussed. For some reason I just am stuck trying to figure out an $r$ which would work.

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3 Answers 3

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Choose arbitrary $(a,b,c)\in S$. Assume without loss of generality that $a,b,c\geq 0$. Put $k=c^2-a^2-b^2-1$.

Then, for any $(a',b',c')\in B((a,b,c),r)$ (with $0<r<c$, $r$ to be specified later -- notice that $c>0$) we have that $$c'^2-a'^2-b'^2-1\geq (c-r)^2-(a+r)^2-(b+r)^2-1=k-2r(a+b+c)-r^2$$ Thus, we can see that if we choose $r>0$ small enough to have $r^2+2r(a+b+c)<k$ (and $r<c$), $B((a,b,c),r)$ will be contained in $S$.

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For the main inequality, do you use $c'^2\ge(c-r)^2$ and so on for each one? I'm trying to see how to get there from having $\sqrt{(a'-a)^2 + (b'-b)^2 + (c'-c)^2}<r$ –  process91 Aug 14 '12 at 1:51
    
@MichaelBoratko: Yes. This is obvious: for $(a',b',c')$ in this ball, $c'$ belongs to the interval $(c-r,c+r)$, and of course $(c-r)^2$ is smaller than the square of any number in that interval (because they're all positive). Similarly for $a',b'$. –  tomasz Aug 14 '12 at 2:00
    
Should we also have $r<a$ and $r<b$ to be sure that $a'$ and $b'$ are positive? –  process91 Aug 14 '12 at 2:06
    
@MichaelBoratko: You can do that, but that is not necessary. If $a\geq 0$, then $(a+r)^2$ is greater than the square of any number in the interval $(a-r,a+r)$, whether or not there are negative numbers within. –  tomasz Aug 14 '12 at 2:09
    
OK, good deal. Thank you very much! –  process91 Aug 14 '12 at 2:19

Note that $|x-x'| \leq \|(x,y,z)-(x',y',z')\|$. Furthermore, we have $|x^2-(x')^2| = |x-x'| |x+x'|$. Hence if $|x|, |x'| < L$, then $|x^2-(x')^2| \leq 2 L \|(x,y,z)-(x',y',z')\|$. The same applies to $y,y', z, z'$, of course.

Pick $(x,y,z) \in S$, let $M = \max\{|x|,|y|,|z|\}$ and $r = \min \{M, \frac{1}{6M}(z^2-(1+x^2+y^2))\}$ (which is $ >0$, of course).

Suppose $\|(x,y,z)-(x',y',z')\| < r$. Then we have $|x^2-(x')^2| < 2M r$, and similarly for the other coordinates.

Finally, we have $(z')^2-(1+(x')^2+(y')^2) > z^2-2Mr - (1 +x^2+y^2) - 4Mr$. By choice of $r$, we have $z^2 - (1 +x^2+y^2) \geq 6Mr$, from which we obtain $(z')^2-(1+(x')^2+(y')^2) > 0$, hence $(x',y',z') \in S$. It follows that $S$ is open since $B((x,y,z), r) \subset S$.

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Oops - I had the inequality going the wrong way in my question - I will fix it –  process91 Aug 14 '12 at 1:28
    
My repair was interrupted by dinner! –  copper.hat Aug 14 '12 at 3:04

Let $f:\mathbb{R}^3 \rightarrow \mathbb{R}$ be the function $f(x, y, z) = z^2 - x^2 - y^2 - 1$. Since $f$ is continuous and $(0, \infty)$ is open in $\mathbb{R}$, $S = f^{-1}((0, \infty))$ is open.

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He wrote that he it not using anything topological. –  copper.hat Aug 14 '12 at 1:19
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I do not have the theorem to back this up yet - only what I put above. Although I suppose I could rip into the details of this in order to reach that conclusion, I doubt that is what was intended. –  process91 Aug 14 '12 at 1:19
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A pity the OP can't use this since it is, by far, the easiest solution, though I believe it must be done with $\,(0,\infty)\,$ –  DonAntonio Aug 14 '12 at 2:33
    
@DonAntonio Someone edited the question. Originally it was $(-\infty, 0)$. –  Makoto Kato Aug 14 '12 at 3:36
    
Indeed @MakotoKato . –  DonAntonio Aug 14 '12 at 3:41

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