Hello,
I have two questions about equivalence in topology.
1) Are an oval and a two dimensional double-torus equivalent?
2) Are a circle and a two dimensional donut equivalent?
Thanks,
MadCow
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Hello,
I have two questions about equivalence in topology.
1) Are an oval and a two dimensional double-torus equivalent?
2) Are a circle and a two dimensional donut equivalent?
Thanks,
MadCow
i think that will depend on how you project the torus/donut into 2-space. Either way, id vote "no"
Good, that was the answer I hope hoping for. Actually I messed up the second question. I meant to ask
3) if a circle and a square are equalivalent.
and
4) if a pentagon and a square are equalivalent.
The premise of the first two questions was that one object contained a hole while the other didn't. I was wondering if the hole could be "closed" by enough continous stretching. I had favored no as eventually you would have to combine combine the stretched out part with the edge of hole, which I questioning the validity of that operation.
The issue with the third and fourth questions is that there is a variance in side numbers. I am thinking no for question 3 and yes for question 4, but someone clearing that up for me would be very helpful.
Thanks again,
MadCow
i dont think i quite understand anymore...
Here, I will rephrase my questions and add a visualization at the end.
1) Are a rectangle and a rectangle with a hole equalivant?
2) Are an octagon and a rectangle equivilant?
3) Are a circle and a rectangle rectangle equivilant?
I hope that's understandable.
Thanks,
MadCow
well, that all depends on how you define "equivalent". Is there more information to these questions? it all seems a little vague
By equivalent I meant one could be tranformed into the other using standard "rubber sheet geometry" rules like you can stretch but not tear, twist, etc. Apart from that I don't have any more information. If that isn't enough I suppose I'll track down some people from school who might be able to help me.
ah ok. Then id say 1 is false, 2 is true, and 3 is false... but i guess the real question is why
Yum...topology. I always wondered why I took that class. If I remember right, the only property that can make two shapes topologically non-equivalent is if two arbitrarily close points from shape A are not arbitrarily close in shape B. In your examples, I'd say 1) false, 2) true, and 3) true.
If I did your homework for you, then you might pass your class without learning how to write a program like this. Then you might graduate and get your degree without learning how to write a program like this. You might become a professional programmer without knowing how to write a program like this. Someday you might work on a project with me without knowing how to write a program like this. Then I would have to do you serious bodily harm. - Jack Klein
1) NoOriginally Posted by Mad Cow
2) Yes
3) Yes
Two sets A and B are equivalent in this sense if there exists a bijective continuous function from A to B.Originally Posted by Perspective
The proof that a circle and a rectangle are topologically equivalent can be done by observing thatOriginally Posted by Perspective
x = r cos t , 0<r<1
y = r sin t , 0<t<2pi
is a circle and the (r,t) space is a rectangle.
Last edited by Sang-drax : Tomorrow at 02:21 AM. Reason: Time travelling