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Math Help - Why is this set open?

  1. #1
    CSM
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    Why is this set open?

    Let X be a completely regular space. And subsets K ⊂ X compact and C ⊂ X closed, with K ∩ C = ∅.
    (i) Show that there are functions fx ∈ C(X, [0, 1]) with fx(x) = 1 and fx = 0 on C, where x ∈ K.
    (ii)Use the functions fx to make an open cover of K
    (iii)Use compactness to make a function g ∈ C(X, [0,∞)) which is g = 0 on C and g > 1/2 on K.
    (iv) Change g to get a f ∈ C(X, [0, 1]) with f = 1 on K and f = 0 on C.
    NOTE: X doesn't have to be normal (T4)!

    ANSWERS:
    X is completely regular and C ⊂ X is closed so for every x not in C there is a fx
    C(X, [0, 1]) with f(x) = 1 and f = 0 on C. We do this for every x ∈ K and define U_x =f_x^{-1}((\frac{1}{2}, 1])\subset X. Ux is open and contains x. So {Ux | x ∈ K} is an open cover of K. K is compact, so there are x1, . . . , xn ∈ K such that Ux1 ∪ ∪ Uxn ⊃ K. We define g(x) = fx1(x) + + fxn(x) or g(x) = max{fx1(x), . . . , fxn(x)}. For both definitions it follows that g ∈ C(X, [0,∞)) and g = 0 on C. Suppose x ∈ K. Then x ∈ Ui for some i, so fi(x) > 1/2.
    So g(x) > 1/2. Let h ∈ C([0,∞), [0, 1]) be such that h(0) = 0 and h(t) = 1 for t ≥ 1/2. Then f = h ◦ g has the desired properties.


    Question: Why is Ux open? What are the other possibilities of making an open cover of K by means of using the fx? Is this the most natural choice?

    T_es^t
    Last edited by CSM; May 6th 2011 at 03:29 AM. Reason: tex
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  2. #2
    CSM
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    Quote Originally Posted by CSM View Post
    Let X be a completely regular space. And subsets K ⊂ X compact and C ⊂ X closed, with K ∩ C = ∅.
    (i) Show that there are functions fx ∈ C(X, [0, 1]) with fx(x) = 1 and fx = 0 on C, where x ∈ K.
    (ii)Use the functions fx to make an open cover of K
    (iii)Use compactness to make a function g ∈ C(X, [0,∞)) which is g = 0 on C and g > 1/2 on K.
    (iv) Change g to get a f ∈ C(X, [0, 1]) with f = 1 on K and f = 0 on C.
    NOTE: X doesn't have to be normal (T4)!

    ANSWERS:
    X is completely regular and C ⊂ X is closed so for every x not in C there is a fx
    C(X, [0, 1]) with f(x) = 1 and f = 0 on C. We do this for every x ∈ K and define U_x =f_x^{−1}((\frac{1}{2}, 1])\subset X. Ux is open and contains x. So {Ux | x ∈ K} is an open cover of K. K is compact, so there are x1, . . . , xn ∈ K such that Ux1 ∪ ∪ Uxn ⊃ K. We define g(x) = fx1(x) + + fxn(x) or g(x) = max{fx1(x), . . . , fxn(x)}. For both definitions it follows that g ∈ C(X, [0,∞)) and g = 0 on C. Suppose x ∈ K. Then x ∈ Ui for some i, so fi(x) > 1/2.
    So g(x) > 1/2. Let h ∈ C([0,∞), [0, 1]) be such that h(0) = 0 and h(t) = 1 for t ≥ 1/2. Then f = h ◦ g has the desired properties.


    Question: Why is Ux open? What are the other possibilities of making an open cover of K by means of using the fx? Is this the most natural choice?

    T_es^t
    Does the Latex work?!
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  3. #3
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    Quote Originally Posted by CSM View Post
    Does the Latex work?!
    If you use [tex]... [/tex] tags.
    [tex] \left\lfloor x \right\rfloor \;\& \,\left\lceil x \right\rceil [/tex] gives  \left\lfloor x \right\rfloor \;\& \,\left\lceil x \right\rceil
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by CSM View Post
    Let X be a completely regular space. And subsets K ⊂ X compact and C ⊂ X closed, with K ∩ C = ∅.
    (i) Show that there are functions fx ∈ C(X, [0, 1]) with fx(x) = 1 and fx = 0 on C, where x ∈ K.
    (ii)Use the functions fx to make an open cover of K
    (iii)Use compactness to make a function g ∈ C(X, [0,∞)) which is g = 0 on C and g > 1/2 on K.
    (iv) Change g to get a f ∈ C(X, [0, 1]) with f = 1 on K and f = 0 on C.
    NOTE: X doesn't have to be normal (T4)!

    ANSWERS:
    X is completely regular and C ⊂ X is closed so for every x not in C there is a fx
    C(X, [0, 1]) with f(x) = 1 and f = 0 on C. We do this for every x ∈ K and define U_x =f_x^{−1}((\frac{1}{2}, 1])\subset X. Ux is open and contains x. So {Ux | x ∈ K} is an open cover of K. K is compact, so there are x1, . . . , xn ∈ K such that Ux1 ∪ ∪ Uxn ⊃ K. We define g(x) = fx1(x) + + fxn(x) or g(x) = max{fx1(x), . . . , fxn(x)}. For both definitions it follows that g ∈ C(X, [0,∞)) and g = 0 on C. Suppose x ∈ K. Then x ∈ Ui for some i, so fi(x) > 1/2.
    So g(x) > 1/2. Let h ∈ C([0,∞), [0, 1]) be such that h(0) = 0 and h(t) = 1 for t ≥ 1/2. Then f = h ◦ g has the desired properties.


    Question: Why is Ux open? What are the other possibilities of making an open cover of K by means of using the fx? Is this the most natural choice?

    T_es^t
    Friend, I tried to figure out how you defined U_x but the code is indecipherable. Could you possibly use the tex commands?
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  5. #5
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    Like D28, I cannot read the OP either.
    I tried to edit the LaTeX, but you changed my edit.
    I think that it is asking about Ursohn's Lemma..
    Maybe you can find that webpage useful.
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  6. #6
    Senior Member Tinyboss's Avatar
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    So Urysohn's Lemma says that in a normal space, every pair of disjoint closed sets can be separated with a function. This problem wants you to show that, if we only require the space to be completely regular, rather than normal, we can still get the result when one set is closed and the other is compact. (An example of the metatheorem that compact sets are like "fat points"--if we can separate a point from a closed set, then we can separate a compact set from a closed set.)

    As you said, we get the f_x's straight from the definition of completely normal. I suspect your definition is U_x=\{f_x^{-1}(1/2,1]\}, which is indeed an open set containing x and missing C. Pick a finite set \{x_1,\dots,x_k\} of points in K such that \bigcup_{i=1}^kU_{x_i}\supset K (by compactness). Then just define F(x)=\sum_{i=1}^kf_{x_i}}(x), which is finite, zero on C, and at least 1/2 on K. The last transformation is trivial.
    Last edited by Tinyboss; May 5th 2011 at 06:13 PM.
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  7. #7
    CSM
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    Okay thanks for your replies, the latex is fixed now. Your definition of the set is correct. My question now is, why is U_x open and why choose that construction of U_x, is it the most straightforward choice? Is it trivial? Enlighten me please.
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  8. #8
    Senior Member Tinyboss's Avatar
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    U_x is open because it's the inverse image of the open set (1/2,1] (remember, it's a subset of the range [0,1], so it is indeed open) under f_x, which is continuous (we're guaranteed that by complete regularity of the space).
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  9. #9
    CSM
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    Thanks. Looking back it is pretty easy, thanks to your input. I took the inverse of [1] which obviously isn't open. So everything went wrong from there on.
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