Prove that a closed interval [0,1] is nonhomogeneous

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Homework Statement


Prove that the closed interval [0,1] is not a homogeneous topology by showing that there's no bijective, open and continuous (bi-continuous) mapping h: [0,1]→[0,1] such that h(1/2)=0.


Homework Equations


The closed interval is equipped with the usually metric. If the mapping mentioned above exists for any x, y such that h(x)=y, then the topology is called homogeneous.


The Attempt at a Solution


The book says it is easy to verify this; I can kind of feel that it doesn't exist, since if h(1/2)=0 then h(1/2+)=h(1/2-) which violates the bijection, but this is obviously not a proof. Any suggestions?
 

Answers and Replies

  • #2
Office_Shredder
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If h(1/2)=0 then a small neighborhood around 1/2 and a small neighborhood around 0 have to be homeomorphic. I.e. we have a homeomorphism between an interval of the form [a,b) and one of the form (c,d). Can you prove that these are not homeomorphic?
 
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I really have no idea. I'll need to assume such homeomorphism mapping exists and then come up with a contradiction, but with no luck. I'm at earlier stage of my study, I don't know what compactness is.
 
  • #4
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Here is my attempt: suppose h exists, h(1/2)=0, let h(1/2-ε)=a>0, h(1/2+ε)=b>0, suppose a<b (a≠b from bijectivity), then by continuity and bijectivity of h there exists x in (1/2-ε,1/2), such that h(x)=a/2, similarly there exists y in (1/2,1/2+ε) such that h(y)=a/2=h(x), where x≠y.

Somehow there seem to be no use of the continuity of h^{-1}, maybe I'm missing something? e.g, the existence of x and y may be less trivial than it looks?
 
  • #5
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Here is my attempt: suppose h exists, h(1/2)=0, let h(1/2-ε)=a>0, h(1/2+ε)=b>0, suppose a<b (a≠b from bijectivity), then by continuity and bijectivity of h there exists x in (1/2-ε,1/2), such that h(x)=a/2, similarly there exists y in (1/2,1/2+ε) such that h(y)=a/2=h(x), where x≠y.

Somehow there seem to be no use of the continuity of h^{-1}, maybe I'm missing something? e.g, the existence of x and y may be less trivial than it looks?
It turns out that the intermediate value theorem I used requires the completeness of real numbers, thus cannot be argued using elementary topological arguments.
 
  • #6
Dick
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It turns out that the intermediate value theorem I used requires the completeness of real numbers, thus cannot be argued using elementary topological arguments.
Ok, you don't know compact. Do you know connected? That's a topological property. You can use that.
 
  • #7
HallsofIvy
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"Homogeneous" basically means "the same throughout". The two points 0 and 1 have the property that there are points of the set only on one side while every other point has other points of the set on both sides. That's the idea of trying to map 1 to 1/2. That can't be done with bi-continuous function precisely for the reason that 1 has only those points that are less in the set while 1/2 has points on both sides.
 
  • #8
Dick
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"Homogeneous" basically means "the same throughout". The two points 0 and 1 have the property that there are points of the set only on one side while every other point has other points of the set on both sides. That's the idea of trying to map 1 to 1/2. That can't be done with bi-continuous function precisely for the reason that 1 has only those points that are less in the set while 1/2 has points on both sides.
That's pretty superficial. Having "points on one side" is not a topological property. This is a topology problem, not a real analysis problem as sunjin09 has already very correctly figured out. You have to abstract the notion of "having points on one side" to something that makes sense in topology. I'm suggesting connectedness.
 
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