Are these two shapes topologically same


by n.karthick
Tags: shapes, topologically
n.karthick
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#1
Aug21-10, 02:58 AM
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Consider the two shapes circle and circle with a line attached it. (Figure is attached) Are they topologically same?. As far as I know if we bend, stretch a shape and attain another shape both are topologically same.

I feel that circle can be stretched in some way and can be made to a circle with line. So both should be topologically same. But text book says both are not.

I am confused. Can anyone help?
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quasar987
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#2
Aug21-10, 03:29 AM
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You can bend and you can stretch, but you cannot break. How do you get from the circle to the circle with line without breaking?
Landau
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#3
Aug21-10, 06:31 AM
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They are homotopy equivalent but not homeomorphic.

n.karthick
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#4
Aug21-10, 09:56 AM
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Are these two shapes topologically same


Quote Quote by quasar987 View Post
You can bend and you can stretch, but you cannot break. How do you get from the circle to the circle with line without breaking?
Ohh i missed something. Circle should be thought of line with infinitesimal width right?. Otherwise it will be a doughnut!!!
So a doughnut and circle with line are topologically same Am i right?

I am new to this. I dont have any idea of homotopy or homeomorphic. Many many new terms are coming here.

Further I dont find still any textbook which could bring differences between various spaces with examples for easy understanding. ( topological spaces, metric space, normed vector spaces and inner product spaces). Mostly mathematical definitions can be find everywhere. What i like to see is teaching through examples. For example say something which belong to normed vector space and not metric space and which belong to normed vector product space and not inner product space . Such examples will be of immense help in understanding easily the classifications of spaces and their mathematical definitions.
Can anyone suggest some learning material for understanding topological spaces for a beginner like me?
quasar987
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#5
Aug21-10, 09:12 PM
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Yes, a circle has no width.

A doughnut and a doughnut with line (not a circle with line) are topologically the same (if you consider that the "line" too has some width.

As a general rule, something that has 1 dimension (i.e. something that has just length) and something that has more than 1 dimension (ex: something that has length and width) are not topologically the same.

One way to see this is imagine that you have a doughnut made out of silly putty. If it is topologically the same as a circle, then you should be able to maneuver (stretch and bend, but not break) the silly doughnut and deform it into a circle, which is clearly impossible, since objects of 1 dimension do not exist in our world.

Two objects are said to be homeomorphic if they are topologically equivalent. It is a synonym.
zhentil
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#6
Sep15-10, 09:59 AM
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Try deleting any point from the circle. Is it connected? Is the same true for the second figure?
lavinia
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#7
Sep15-10, 10:51 AM
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There is no part of the circle that is homeomorphic to the cross point of the line and circle. But you need to prove this.
lavinia
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#8
Sep15-10, 11:02 AM
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the idea of "same" in topology has many meanings. Each refers to an equivalence relation. You seemed to be referring to the equivalence relation of homeomorphism but there are others. For instance two spaces may be considered the same if they have the same homotopy type or if one is a strong deformation retract of the other. each of these relations isolates topological properties that the spaces have in common and allows analysis of these properties while avoiding the other properties of the spaces.
cesiumfrog
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#9
Sep15-10, 07:48 PM
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Topological isomorphisms should preserve the number of distinct approach trajectories that any point has..
zhentil
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#10
Sep15-10, 09:21 PM
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Quote Quote by cesiumfrog View Post
Topological isomorphisms should preserve the number of distinct approach trajectories that any point has..
I'd be interested to see a mathematical definition of that last part that holds for any topological space.
cesiumfrog
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#11
Sep15-10, 11:17 PM
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Maybe I'm wrong. But for any isomorphism, there will be a specific one-to-one identification of points in one space to points in the other. Preserving topology means statements about open sets will be unaffected. Perhaps you would define a trajectory as any open set that has the given point on its boundary (i.e., every trajectory includes points from any open set containing the point, but no trajectory contains the point). Now, at the three way intersection point, we can easily find three nonoverlapping sets such that: each trajectory overlaps one of these three, and each of the three has a trajectory that overlaps neither of the others. But for any point on a line or circumference a contradictory property holds (ie. the same property except with pairs of sets, never triplets), so they cannot be topologically equivalent.


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