1. Limited time only! Sign up for a free 30min personal 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!

Rudin Problem 2.7

  1. Oct 2, 2014 #1
    1. The problem statement, all variables and given/known data

    Let [itex] A_1, A_2, A_2,...[/itex] be subsets of a metric space.
    (a) If [itex]B_{n}=\bigcup_{i=1}^{n}A_{i} [/itex], prove that [itex] \bar{B_{n}}= \bigcup_{i=1}^{n}\bar{A_{i}} [/itex] for n=1,2,3,...
    (b) If [itex]B=\bigcup_{i=1}^{\infty}A_{i} [/itex], prove that [itex] \bar{B} \subset \bigcup_{i=1}^{\infty}\bar{A_{i}} [/itex],
    Show by example this inclusion is proper.



    2. Relevant equations

    If E is a subset of a metric space, then the closure of E,
    [itex] \bar{E}=E \bigcup \acute{E} [/itex], where [itex] \acute{E} [/itex] is the set of all limit points of E.

    For a subset E of a metric space X, a point [itex] p \in X [/itex] is a limit point if and only if
    [itex] \forall \epsilon > 0, \exists q \in B(p;\epsilon) \bigcap E : q \neq p [/itex]

    3. The attempt at a solution

    Now I think I've got part (a) and (b) nailed down, but I'm having trouble thinking of an example for a proper subset. Here's what I have thus far...

    Suppose [itex]x \in \bar{B_{n}} [/itex], then [itex] x \in B_{n} [/itex] or x is a limit point for [itex] B_{n} [/itex].
    If [itex]x \in {B_{n}} [/itex], then [itex]x \in \bigcup_{i=1}^{n} A_{i} [/itex], then [itex] \exists i, x \in A_{i} [/itex], so [itex] \exists i, x \in \bar{A_{i}} [/itex].
    If x is a limit point, then
    [itex] \forall \epsilon > 0, \exists y \in B(x;\epsilon) \bigcap B_{n}: y \neq x [/itex], so [itex] \forall \epsilon > 0, \exists y \in B(x;\epsilon) \bigcap (\bigcup_{i=1}^{n} A_{i}): y \neq x [/itex], then
    [itex] \exists i,\forall \epsilon > 0, \exists y \in B(x;\epsilon) \bigcap A_{i}: y \neq x [/itex], then
    [itex] \exists i, x\in \bar{A_{i}}[/itex], so [itex] x \in \bigcup_{i=1}^{n}\bar{A_{i}}[/itex]

    Suppose [itex] x \in \bigcup_{i=1}^{n}\bar{A_{i}}[/itex], then [itex] \exists i, x \in A_{j}[/itex] or for some i x is a limit point.
    If [itex] \exists i, x \in A_{i}[/itex], then [itex] x \in \bigcup_{i=1}^{n}A_{i}=B_{n}[/itex].
    If x is a limit point for some A sub i, then x is a limit point for [itex] \bigcup A_{i}=B_{n} [/itex].

    For part (b), I did the same thing, but replaced [itex] B_{n} [/itex] with [itex] B [/itex] and in the union I replaced n with infinity.

    Now, I'm having some difficulty coming up with an example that would show B is a proper subset of the union of an infinite number of A sub i's.

    Any hints or corrections would be much appreciated.
     
  2. jcsd
  3. Oct 2, 2014 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    I can see a fixable flaw in your argument for (a). For simplicity, take ##n = 2##; that is not a loss of generality, since you can then get the general case of finite ##n## by induction.

    Anyway, say ##x \in \bar{B}_2## is in the boundary of ##B_2## but not in ##B_2## itself. Let ##\epsilon_1 > \epsilon_2 > \cdots \to 0##. For ##\epsilon_i## there is a point ##x_i \in B_2 \cap N(x,\epsilon_i)##. It is conceivable that we could have, for example, ##x_1 \in \ A_1, x_2 \in A_2, x_3 \in A_1, x_4 \in A_2, \ldots## so the sequence ##\{x_i\}## is not in a single set ##A_1## or ##A_2##, but, instead, oscillates back and forth between the two. In that case you would need to do more work to conclude that ##x## is in ##\bar{A}_1 \cup \bar{A}_2##. It can be done, but some extra steps are needed; I think the fact that you are working an a metric space matters a lot here.
     
  4. Oct 2, 2014 #3

    pasmith

    User Avatar
    Homework Helper

    Are you sure the inclusion in part (b) is the right way round? Certainly one can show by example that [tex]
    \bigcup_{i=1}^\infty \bar{A}_i \subsetneq \bar{B} [/tex] by taking a bijection [itex]f : \{1, 2, 3, \dots \} \to \mathbb{Q}[/itex] (such a bijection exists because the rationals are countable) and setting [itex]A_i = \{f(i)\} \subset \mathbb{R}[/itex].
     
  5. Oct 2, 2014 #4
    So what you're saying is that just because [itex] \forall \epsilon >0,\exists y \in N(x;\epsilon) \bigcap \bigcup_{i=1}^{n} A_{i} [/itex], I can't conclude that [itex] \exists i, \forall \epsilon >0,\exists y \in N(x;\epsilon) \bigcap A_{i} [/itex]?

    Would I have to say that for a sequence of epsilons, there is a y in the intersection of the Neighborhood with center x and radius epsilon? I thought about distributing the intersection/union, but that doesn't make much sense to me, intuitively.
     
  6. Oct 2, 2014 #5
    pasmith,

    Wow, really nice example. Am I correct in saying "because each of the singletons have no limit points, the union of the A sub i's are the rationals, but the closure of B is the reals"?

    And yes, I was wrong. It should be....

    Show by an example, that this inclusion [itex] B \supset \bigcup_{i=1}^{\infty} A_{i} [/itex] can be proper.
     
  7. Oct 2, 2014 #6

    pasmith

    User Avatar
    Homework Helper

    The point is not that a singleton has no limit points, but that it contains its only limit point and is therefore closed, so that [itex]\bigcup A_i = \bigcup \bar{A}_i[/itex]. Otherwise you are correct, although you need to explain briefly why [itex]\bigcup_{i=1}^\infty \{f(i)\} = \mathbb{Q}[/itex].
     
    Last edited: Oct 2, 2014
  8. Oct 2, 2014 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You probably can conclude it, but it does not follow right away from what you did; you need some extra steps.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Rudin Problem 2.7
  1. 2 Rudin Problems (Replies: 0)

  2. Baby Rudin Problem 2.7 (Replies: 2)

  3. Baby Rudin problem (Replies: 4)

Loading...