Timeline for What polygons can be shrunk into themselves?
Current License: CC BY-SA 3.0
17 events
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| Oct 10, 2014 at 5:02 | comment | added | Gabriel C. Drummond-Cole | every shrinking is a dilation centered at a point in the convex hull. Therefore to consider all possible shrinkings it suffices to consider all dilations centered at all points in the convex hull. The argument is intended to show that if the polygon is not star-shaped, then non-trivial dilations centered at points in the convex hull have to either leave the original polygon or have dilation factor bounded away from one. It seems like both you and Włodzimierz have problems with my answer but I still don't understand them. Please continue to elaborate and hopefully I will see! | |
| Oct 10, 2014 at 2:08 | comment | added | Boris Bukh | @GabrielC.Drummond-Cole, that does not address my question. You start with a single non-trivial shrinking, and consider its powers, get a fixed point of that shrinking, and then consider dilations by factors completely unrelated to the shrinking you started. There is no a priori reason why those would actually be inside the polygon. (Again, the more detailed solution below does address this issue, but not yours.) | |
| Oct 8, 2014 at 16:36 | comment | added | Wolfgang | Note how this can be counter-intuitional: Imagine a half unit disc from which we remove a half disc with radius $\epsilon$ and the same center. The result is a U-shape which is not star-shaped. This is obvious as we let $\epsilon$ grow close to $1$, but hard to imagine for tiny $\epsilon$, yet not shrinkable! | |
| Oct 8, 2014 at 1:04 | comment | added | Gabriel C. Drummond-Cole | if $f(x) = ax + b$ has a fixed point $z$, then $f(z+c) = z + ac$ so $f$ is a dilation centered at $z$. | |
| Oct 7, 2014 at 14:33 | comment | added | Boris Bukh | I do not follow; why does the existence of a fixed point imply that we need to consider only the dilations centered at $x$? (It is clear in the proof of Włodzimierz Holsztyński, but not in this proof. Is there a transparent way of seeing it?) | |
| Oct 7, 2014 at 13:46 | comment | added | Gabriel C. Drummond-Cole | If you believe that each epsilon is positive and continuous then delta is positive by compactness. Do you have a problem with one or both of those? | |
| Oct 7, 2014 at 13:44 | comment | added | S. Carnahan♦ | @WłodzimierzHolsztyński Presumably, you need to show that the supremum function $\epsilon(x)$ is lower semicontinuous with respect to $x$, but this follows from the fact that you can use the same bad point $y$ for points near $x$ with negligible decrease in $\epsilon$. | |
| Oct 7, 2014 at 11:39 | comment | added | Włodzimierz Holsztyński | @Gabriel, perhaps you may help me, and make the presentation of your answer easier to me. I'll perhaps come back later. At this moment let me just mention that you didn't say why your $\ \delta\ $ is positive. (Perhaps it's easy but it still needs at at least a clear comment, it seems to me). | |
| Oct 7, 2014 at 11:22 | comment | added | Gabriel C. Drummond-Cole | Yikes, more than one? I'd like to hear about the problems you see. | |
| Oct 7, 2014 at 5:17 | comment | added | Włodzimierz Holsztyński | The whole considerations about the fixed point (given the Banach's fixed point theorem) are unnecessary. On the top of it I have doubts, more than one, that this proof is valid. | |
| Oct 2, 2014 at 16:37 | comment | added | Erel Segal-Halevi | Thanks. It is gladdening to have a necessary condition that matches the sufficient condition. | |
| Oct 2, 2014 at 11:51 | vote | accept | Erel Segal-Halevi | ||
| Oct 2, 2014 at 9:18 | history | edited | Gabriel C. Drummond-Cole | CC BY-SA 3.0 |
extended the argument to a more general case.
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| Oct 2, 2014 at 9:12 | comment | added | Gabriel C. Drummond-Cole | Right, just temporarily replace $D$ with its convex hull and then you get a fixed point somewhere in the convex hull and the same argument applies whether or not the fixed point is in $D$ itself or not. | |
| Oct 2, 2014 at 9:01 | comment | added | Gabriel C. Drummond-Cole | I was using it for the algebro-topological proof that there is a fixed point. I think the iterative construction (replace $D$ with the image of the shrink map) should work to give a fixed point in the general case. | |
| Oct 2, 2014 at 8:55 | comment | added | Erel Segal-Halevi | Why do you need to assume that the original polygon is simply-connected? | |
| Oct 2, 2014 at 8:46 | history | answered | Gabriel C. Drummond-Cole | CC BY-SA 3.0 |