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Homework Help: 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

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    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


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    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

    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


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    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

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    You probably can conclude it, but it does not follow right away from what you did; you need some extra steps.
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