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Homework Help: Urysohn Lemma

  1. Dec 16, 2005 #1


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    This is a problem I am doing for review, in the section of my book on the Urysohn lemma:

    A [itex]G_{\delta}[/itex] set in a space X is the intersection of a countable collection of open sets in X. Let X be normal. Prove that if A is a closed [itex]G_{\delta}[/itex] set in X, then there exists a continuous function [itex]f : X \to [0, 1][/itex] such that f(x) = 0 if x is in A, and f(x) > 0 otherwise.

    The Urysohn lemma states:

    X is normal, A and B are disjoint closed subsets of X. Let [a,b] be a closed interval in the real line. Then there exists a continuous map

    [tex]f : X \to [a, b][/tex]

    such that f(x) = a for every x in A, and f(x) = b for every x in B.

    The proof for this lemma has some open set U0 containing A such that f(U0) = {0}, and it seems possible that f might be 0 even for some points outside this U0. But I need a function that is 0 on A, and only on A. Now the proof of the lemma also has a bunch of sets Up corresponding to each rational p in [0, 1] such that p > q implies [itex]\overline{U_q} \subset U_p[/itex]. The fact that A is a [itex]G_{\delta}[/itex] set tells me that

    [tex]A = \bigcap _{n = 1} ^{\infty} A_n[/tex]

    I suppose I could also re-write A as:

    [tex]A = \bigcap _{n=1} ^{\infty} \left ( \bigcap _{k = 1} ^{n}A_k\right ) = \bigcap _{n=1} ^{\infty} B_n[/tex]

    so that A is an intersection of a descending chain of open sets. The thing is, the An and/or the Bn are not, a priori, like the Uq in that

    [tex](U_p \subset U_q \wedge U_p \neq U_q) \Rightarrow \overline{U_p} \subset U_q[/tex]

    I'm not sure what to do. I was thinking to first look at A and [itex]B_1^C[/itex] as my first two closed sets, and use the Urysohn lemma to define a continuous function from X to [1/2, 1] choosing [itex]U_{1/2}^1[/itex] to be [itex]B_2[/itex]. Next, look at A and [itex]B_2^C[/itex] as my closed sets, and use the Urysohn lemma to find a function from X to [1/3, 1/2] choosing [itex]U_{1/3}^2[/itex] to be [itex]B_3[/itex]. In case it's not clear, I essentially plan on applying the Urysohn lemma infinitely many times, and the kth time I apply it, the smallest open set containing A that is used in defining the function will is what I'm calling [itex]U_{(k+1)^{-1}}^k[/itex].

    I'll then get a bunch of functions, the kth one mapping Bk+1 to 1/(k+1) and [itex]B_k^C[/itex] to 1/k. I will just look at how these functions behave on [itex]\overline{B_{k} - B_{k+1}}[/itex] and then use the pasting lemma to argue that it is continuous. Does this seem like a workable approach?
    Last edited: Dec 17, 2005
  2. jcsd
  3. Dec 16, 2005 #2


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    There are definitely some problems with the approach so far. I'll try to fix it, and then see if this fix is possible. I also don't think that what I have above is all that clear.

    A is a [itex]G_{\delta}[/itex] set, so it can be written as the countable intersection [itex]\cap A_n[/itex]. Define:

    [tex]B_n = \bigcap _{k=1} ^n A_k[/tex]


    [tex]A = \bigcap _{n = 1} ^{\infty} B_n[/tex]

    (Bn) is a descending sequence of open sets. Pick a subsequence (Cn) such that C1 = B1, and Cn+1 is the largest element of the B sequence contained in Cn such that

    [tex]\overline{C_{n+1}} \subset C_n[/tex]

    (* I don't know yet that this is possible, but we'll see later). Define functions

    [tex]f_n : X \to \left [\frac{1}{n+1},\frac{1}{n}\right ][/tex]

    that are continuous, and such that f(Cn+1) = {(n+1)-1} and f(X - Cn) = {n-1}. Do this by applying the Urysohn lemma to disjoint closed sets A and X - Cn, and picking Cn+1 to be the open set containing A whose closure is contained in X - (X - Cn) (just like the Urysohn lemma, when applied to the closed disjoint sets A, B and the interval [0,1] uses the normality of X to choose an open set U0 containing A whose closure is contained in X - B).

    Consider the functions:

    [tex]g_1 = f_1|_{X - C_2},\ g_2 = f_2|_{\overline{C_2} - C_3},\ \dots ,\ g_n = f_n|_{\overline{C_n} - C_{n+1}},\ \dots[/tex]

    These functions are continuous on closed sets whose union is X. Two of these functions have overlapping domains when their indices are adjacent. In particular, gn and gn+1 have domains that overlap precisely on the boundary of Cn+1 which is

    [tex]\overline{C_{n+1}} - C_{n+1}[/tex]

    This functions will agree on the overlapping portions, so although the pasting lemma is only for two functions defined on closed sets whose union is X, I would like to say that I can apply to this (possibly infinite) countable set of functions, and thereby get a continuous function from all of X to [0, 1] by pasting all these functions together, to get a function f.

    Now f(A) would defined to be {0}, and this would be continuous. Any point outside A would be outside some Cn, and would thus be greater than 0.

    Going back to *, suppose such a choice is not possible. Then there is some Cn = Bm such that for all l > m, the closure of Bl is not contained in Bm. But by normality of X, there is some open subset of X, B, such that B contains A, and Bm contains B's closure. So maybe instead I can choose [itex]C_{n+1} = B_{m+1} \cap B[/itex] and replace the sequence

    [tex]\langle B_l \rangle _{l > m+1}[/tex]


    [tex]\langle B_l \cap B \rangle _{l > m+1}[/tex]

    and repeat this argument to find the sequence <Cn>.
  4. Dec 17, 2005 #3


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    You sure you have the problem right?

    For any continuous function f, the zero set of f must be a closed set. (Because it is the inverse image of the closed set {0}) [itex]G_{\delta}[/itex] contains sets that are not closed, so this can't be true.
  5. Dec 17, 2005 #4


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    I forgot to write that A is a closed [itex]G_{\delta}[/itex] set.
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