1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Urysohn Lemma

  1. Dec 16, 2005 #1

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    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

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    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]

    Then:

    [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]

    with

    [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

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    I forgot to write that A is a closed [itex]G_{\delta}[/itex] set.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Urysohn Lemma
  1. Urysohn's lemma (Replies: 4)

  2. Ito's Lemma (Replies: 0)

  3. Basic Lemma (Replies: 1)

  4. Banach's Lemma (Replies: 1)

Loading...