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Proposed: Quotient Space

  1. Sep 26, 2011 #1
    Let X be the quotient space obtained of [itex]\mathbb{R}[/itex] identifiying every rational number to a point. Is X a Hausdorff space? Is X a compact space?
     
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  3. Sep 26, 2011 #2

    quasar987

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    So the quotient space can be identified as a set with (R\Q) u {P}, where P is the point onto which all the rational numbers collapse. A space is Hausdorff if every pair of points can be "separated by open sets". Take P and any irrational i. Can they be separated by open sets? Hint: recall from real analysis that Q is dense in R.

    For compactness, think about how you could construct a particular open covering of the quotient space that has no finite subcover.
     
  4. Sep 26, 2011 #3
    I think the topology you get is the cofinite , aka, finite complement topology:

    every open set in R/~ must contain the whole of Q, i.e., every rational, since an

    open set in R must contain an interval. The only way your set will be open in the

    quotient topology, if it contains all-but-finitely-many irrationals , then q^-1(S)=

    R\\/ {i_1,..,1_n} ,for i_1,..,i_n irrational, is the complement of the closed set {i_1,..,1_n}.

    This topology is not only not Hausdorff, but it is "anti-Hausdorff " , in that no two elements

    can be separated. I may be missing some open sets, e.g., maybe R- {i_1,..,i_n,...} , i.e.,

    we may be able to miss countably-infinite many elements.
     
  5. Sep 27, 2011 #4

    lavinia

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    A set is open in the quotient space only when its inverse image under the quotient mapping is open. If the quotient were Hausdorff then the inverse image of two points can be separated by open sets. In this case, this is not possible. Consider an open neighborhood of the point that is the projection of the rationals. An open set must pull back to an open neighborhood of every rational and so must be the entire real line. What about an open set around the projection of an irrational?

    This space looks non-compact to me.
     
    Last edited: Sep 27, 2011
  6. Sep 28, 2011 #5

    lavinia

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    Thinking about it more it seems that the space is compact. What do you think?
     
  7. Sep 28, 2011 #6
    I think I was wrong :

    We have R/Q:= { Q, {i}} i is irrational

    (as a set).



    Then p^-1( R/Q-{io} ) =R-{io} is open, so is

    p^-1( R/Q-[{io}\/{ij}] ) =(-00,io)\/ (io,ij)\/ (ij,oo) (assume io<ij ; both are irrational)

    and so on. So I think this determines all the open sets.
     
  8. Sep 29, 2011 #7
    Is it not true that this set has the trivial topology?

    Take an open set in R/Q which is non-empty. It is open if and only if its inverse image in R is open- so for a start we need it to contain a rational number. So let's look at the inverse image of this set. We know it contains a rational so the inverse image will have at least all of the rationals in it. But then, of course, it must contain all of the irrationals too since the only open set in R containing all of the rationals is the whole line.

    Hence, our space is an uncountable set equipped with the trivial topology- the only open sets are the empty set and the whole set. Hence, it is (very!) non-Hausdorff, and (very!) compact.

    And I suppose R/Ir (R mod the irrationals) would be the same story, except that you would have a countable set.
     
  9. Sep 29, 2011 #8
    In fact, isn't this simply true in general? If I have a space X and a dense subspace Y, then X/Y will be a space with the trivial topology?

    I believe that this is the foundational philosophy of noncommutative geometry. Sometimes, for example, it may be interesting to look at the orbits of points in a dynamical system- but it is unhelpful to look at the topology of the quotient space if those orbits are dense since you get something with trivial topology which is practically useless. However, there will be a non-commutative space which is an analogue of the orbit space which will contain very rich information.
     
  10. Sep 29, 2011 #9
    Not, so, Jamma, that the only open subset of R containing all rationals is R itself :

    Take, e.g., {Pi} (i.e., the irrational real number 3.1415927.... ). In the topology of R , singletons/points are closed sets ( I think

    the official name is that R is To ).Another proof (that generalizes to metric spaces)

    that singletons are closed is this:

    consider x in R , and x' in R-{x} ( i.e., x'=/x ). Then d(x,x')=r>0 , so the ball

    (x'-r/2,x'+r/2) is an open ball containing x', and it is contained fully in R-{x},

    this shows R-{x} is open, so R-(R-{x})={x} is closed in R.

    Then R-{Pi} is the complement of a closed set ,

    therefor open. Same goes for the complement in R of two singleton irrationals.
     
    Last edited: Sep 29, 2011
  11. Sep 29, 2011 #10
    Oh yeah, derp!

    So what is true is that the closure of any open set is the whole set R/Q, so the open sets are all dense subsets- not the whole set as I said, sorry about that. So this is non-Hausdorff (no two points can be separated) but not compact, since I could choose my open sets as follows:

    Pick some irrational (pi, say). Define the open sets U_n (n an integer) as the whole of R/Q with the points i.pi removed for integers i>n. These sets are clearly open (their inverse image in R is just a union of open intervals of the form (-\infty,(n+1)pi),((n+1)pi,(n+2)pi),....) and their union covers R/Q but no finite collection will cover it.
     
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