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Can a nontrivial quotient space of R be homeomorphic to R?

  1. Mar 19, 2013 #1
    Intuitively, one would assume that the quotient space of a topological space under an equivalence relation would always be smaller than the original space. It turns out this is not remotely true. I'm specifically interested in quotient spaces of ℝ (under the standard topology).

    We can easily make a quotient space of ℝ be homeomorphic to ℝ, for instance be gluing all the points in an interval into a single point. We can even glue infinitely many intervals into points, and still get a quotient space homeomorphic to ℝ. But my question is this: let us call an equivalence relation "nontrivial" if every equivalence class has at least two elements. Then does there exist a nontrivial equivalence relation on ℝ such that the quotient space is homeomorphic to ℝ?

    Any help would be greatly appreciated.

    Thank You in Advance.
     
  2. jcsd
  3. Mar 20, 2013 #2

    micromass

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    There is a theorem that if ##f:X\rightarrow Y## is a closed (or open) continuous surjection, then it is a quotient map. So if we define for ##x,y\in X## the following equivalence relation

    [tex]x\sim y~\Leftrightarrow ~ f(x)=f(y)[/tex]

    then we have that ##Y=X/\sim##.

    So if we succeed to find a closed (or open) continuous surjection ##f:\mathbb{R}\rightarrow \mathbb{R}##, then we will have found an equivalence relation such that its quptient is ##\mathbb{R}##.

    I think that the following

    [tex]f:\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow x\sin(x)[/tex]

    is a closed and continuous surjection. So this would be an example.
     
  4. Apr 2, 2013 #3
    Thanks micromass. What if we imposed an additional condition on the equivalence relation: if a<b<c and a~c, then a~b. Under that condition, it's impossible to have the quotient space be homeomorphic to ℝ, right?
     
  5. Apr 4, 2013 #4

    micromass

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    Yeah, I think it should impossible then. But let me think of a proof...
     
  6. Apr 5, 2013 #5
    With the additional constraint, the set of image points with more than one origin cannot be uncountable, since each preimage contains an open interval. From separability of [itex]\mathbb{R}[/itex] there is no uncountable set of pairwise disjoint open intervals.
     
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