Just to clarify,is the three-sphere the only type of bounded three-dimensional space possible that contains no holes?
2006-08-20 11:16:35 · 5 answers · asked by redlens 3 in Science & Mathematics ➔ Mathematics
Poincare conjecture: stupid...
Some dude ate a donut once and asked a stupid question once and it took mathematicians 10 years to answer it...stupid bast*rds
2006-08-20 23:17:50 · answer #1 · answered by Anonymous · 0⤊ 1⤋
This is the Poincare conjecture and is a Millenium problem. It seems that it was solved in outline by Perleman and in detail by a couple of Chinese mathematician. The specific proof was actually of the geometrization conjecture, which was known to imply the Poincare conjecture.
So the answer to your question is YES.
It seems that Doug is uncharacteristically confused by this subject. Yes, homeomorphism does preserve simple connectivity, but that is not the question. Also, it is quite easy to discuss homeomorphism between spaces that are neither compact nor Hausdorff. It is known that a continuous bijection from a compact space to a Hausdorff one must be a homeomorphism (which is what I think the confusion is about), but this is only one specific result among a number concerning these concepts.
2006-08-20 13:20:53 · answer #2 · answered by mathematician 7 · 2⤊ 0⤋
The answer is yes because connectivity has to be preserved under homeomorphism (by definition) which means that any line C which is finite and encloses an area A has lim A -> 0 C ->0 implying that there are no 'holes' to draw the line 'around'.
Homeomorphism is a somewhat stronger requirement on
f:X -> Y than is simple bijection since it's fairly easy to show that for f to be a homeomorphism, X must be a compact space and Y must be a Hausdorf space.
Doug
2006-08-20 11:54:00 · answer #3 · answered by doug_donaghue 7 · 1⤊ 1⤋
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2016-11-26 20:18:56 · answer #4 · answered by mitra 4 · 0⤊ 0⤋
Ummm, yeah....
2006-08-20 12:29:00 · answer #5 · answered by jennifer c 3 · 0⤊ 3⤋