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Let me state my problem. Suppose we have a ball $B$ in standard $\mathbb{R}^3$, that is a $\varepsilon$-neighbourhood of $0$ point. Suppose we have a family of cones $X_C = \lbrace C > 0 \vert x^2 + y^2 \leqslant C \cdot z^2 \rbrace $. Also we have a homeomorphism $h$ that maps $B$ on itself ($h(B) = B$) and $h(0) = 0$ (if it's crucial, $h(B \cap Oz) = B \cap Oz$ and $h(0) = 0$). So, the question is: if we take a cone, corresponding to some value $C_1$, does exist some $C_2$ that $f(X_{C_1}) \subseteq X_{C_2}$?

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Does $Oz$ mean the $z$-axis intersected with the ball $B$? – Lee Mosher Jul 11 '12 at 16:35
Whether it does or not, though, this questions seems to me to be more suited to other sites such as It is not a research level question, which is what Math Overflow is for, as you will see if you read the faq. – Lee Mosher Jul 11 '12 at 16:59
Yes, I meant that. Sorry, I'll correct this moment – Evgeny Jul 11 '12 at 17:00
That question appeared in my research, so I just wanted to check is this a common fact (that I've missed in my education) or should I cope with that question myself. – Evgeny Jul 11 '12 at 17:05
I guess you meant $h(X_{C^*}) \subseteq X_{C^*}$! – Mercy King Jul 11 '12 at 17:31

1 Answer 1

up vote 1 down vote accepted

Here is the counterexample, re-expressed. Let $S_r$ be the sphere of radius $r \epsilon$. Construct a homeomorphism $h : B \to B$ with the following properties:

  • $h(S_r) = S_r$ for each $r \in [0,1]$
  • $h(S_{1/3} \cap X_{C^*}) = S_{1/3} \cap X_{C^*}$
  • $h(S_{2/3} \cap X_{C^*}) \supset S_{2/3} \cap (\mathbb{R}^3 - X_{C^*})$
  • And, if you like, $h$ preserves the $z$-axis, which prevents the last $\supset$ from being $=$.
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Well, thanks for idea, I've constructed one example of that kind and my question has negative answer there; but I couldn't preserve $z$-axis. So far, I've discovered another example by myself that preserves $z$-axis, but too has negative answer. I just took upper half of cone and laid it on disc $B \cap Oxy$ (and constructed images for other parts of ball correspondingly). So, thank you for idea that my question doesn't have positive answer in all cases! – Evgeny Jul 11 '12 at 20:30

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