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Urysohn lemma?

  1. Jul 13, 2006 #1
    It roughly says there exists a continous function from a normal space X to some interval [a,b]

    Since the the space is a normal space, there exist two disjoints closed subsets A and B.

    What I don't understand is how can you associate some abstract space with a real interval and is continous as well?
     
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  3. Jul 13, 2006 #2

    HallsofIvy

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    I'm afraid you'll have to clarify your question. Certainly you can define a function from any set to any other. Since, in this case, both domain and range sets are topological spaces, continuity is also defined.
     
  4. Jul 13, 2006 #3
    what I don't get is how you can have a continous tranformation from disjoint sets to a continous interval in R.
     
  5. Jul 13, 2006 #4

    matt grime

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    Why don't you get that? Consider the subsets [0,1) and [1,2], they are disjoint and the obvious map is continuous map onto the interval [0,2].

    However, this doesn't really bare much relation to the Urysohn lemma which staes that in a normal space, S, given two disjoint (open) sets A and B there is continuous map f from S to [0,1] with f(A)=0 f(B)=1.

    How you should think of this is as follows: we'll define f(A)=0, now, we'll put A inside lots of nested subsets that you should think of as filling out to occupy all of S and in such a way that f(B)=1
     
  6. Jul 13, 2006 #5

    HallsofIvy

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    For example: Let A= (0, 1), B= (2,3). Then the "Urysohn function" might be f(x)= 0 if 0<x < 1, f(x)= x-1 if [itex]1\le x\le 2[/itex], f(x)= 1 if 2< x< 3.
     
  7. Jul 13, 2006 #6
    Thanks for the insights.

    Just a have another quick question, Is such a map a homemorphism?
     
  8. Jul 14, 2006 #7

    matt grime

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    How can it be? it has to be constant on A and B so if they're not one point setes then it fails to be bijective straight away,

    Further more you're defining a map to the interval [0,1]. Are you saying you think all normal spaces are homeomorphic to the unit interval?
     
  9. Jul 14, 2006 #8

    mathwonk

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    look at it backwards. if f is a continuous function from a space X to the interval [a,b], then the inverse image of {a} is clsoed,a s is the inverse image of {b}.

    what about a converse? i.e. given closed sets A, B in X, is there a continuous function f on X such that they are the inverse images of {a} and {b}? urysohn says what you need to assume abut X for that to be true.
     
  10. Jul 14, 2006 #9
    "given closed sets A, B in X, is there a continuous function f on X such that they are the inverse images of {a} and {b}"

    That makes alot more sense now. I always like to think that everything is homemorphic, but I fail see that.

    I'm a visual learner, so I'm trying to visual this as blobs of areas mapping to an interval. But I'm not sure if thats a correct interpratation either.

    I'm studying the proof now, it's a killer, heh.
     
  11. Jul 14, 2006 #10

    mathwonk

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    naw its trivial, just tghink backwards, i.e. how do you define afunction from its levels ets/ i.e. a function isd ewtermiend if you know for all real t where it equals t.

    so just set it equal to 0 on A and equal to 1 on B. then where should it equal 1/2? on some set that separates A from B. lets see, i guess you just need to construct a sequence of closed sets.


    hmmm what is a nokormal space anyway? you can separate any two clsoed sets by open sets? or soemthing?

    anyway given A and B choose two more clsoed sets, disjoint and havcing A,B respectivel;y in their interiors. These will be where the function is <= 1/4 and >= 3/4 or something. then do it again,...

    its a littl complicated but just work backwards, i.e. think of having sucha fucntion and ask what the sets look, olike where it is <= k/2^n for all k,n. then the definition fo normals epace is amde so you can actually have such sets nesated in the right way, and them you just define the fucntion to behave as it should.

    so tyuo should try it first then read it then try to understand it this way, or any way you like. but it is trivial. just complicated.
     
  12. Jul 15, 2006 #11

    matt grime

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    I tend to think of Urysohn in terms of balloons. This is just a visual way of relating mathwonk's idea.

    Normal means (I tihnk) that closed sets are separated, so imagine an open set around A as a balloon and you inflate it until it touches B. Think of time parametrizing the inflation as your function from [0,1].
     
    Last edited: Jul 15, 2006
  13. Jul 15, 2006 #12

    mathwonk

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    i like that makes it seem simpler and more imaginable.
     
  14. Jul 16, 2006 #13
    Havent had a chance to look over the proof yet, but now I'm giving it a stab The inflating ballon example makes alot more sense. The more I think about it, the more trivial it is indeed. But the details are still a little murky. For instance, the open sets enclosing A that are inflated as time goes on, are indexed to rational numbers. This is from Munkres book. The way I see it, time parametrizing skips over all irrationals on [0,1]?
     
  15. Jul 16, 2006 #14

    matt grime

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    It won't skip the irrationals at all. (If it skipped any value, t, in [0,1] then the orginal set must be disconnected into at least two open/closed components, the inverse images of [0,t) and (t,1], and [0,t], [t,1], which are open/closed in the subspace topology. Thus we have at least two components one around each of A and B making the construction of such a function trivial indeed: make it 0 on the one containing A and zero on the one containing B, and do whatever you want on the other components.)

    The point about continuiuty is that loosely you can infer values from surrounding points, can't you?
     
    Last edited: Jul 16, 2006
  16. Jul 17, 2006 #15

    mathwonk

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    this is basic to reals. all can b e described as limits of sequences of rationals. the countability of the rationals, makes the construction feasible. then you take a limit to get the irrationals.
     
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