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| | Equivalence of the Box/Ball topology - Vectorcalcumb |
 | | So, when we say that the ball and box topology are equivalent, we mean that for any neighbhorhood in either, there is a neighbhorhood in the other which contains the original neighbhorhood, right? |
 | | On going to write just now, I had thought that this was clearly impossible, since you can't square the circle, but apparently you can square the circle, just not with a ruler and compass(Which, of course makes perfect sense, it's just a square with |
 | | However, it still seems to me that even if we have a neighbhoorhood of equal area(volume, extent) around a point, we can't possibly have an elementwise identical neighbhoorhood in the other topology, just one defined to have the same extent in the other topology. |
| editthis.info /VectorCalcUMB/Equivalence_of_the_Box/Ball_topology (164 words) |
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