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Intersection of a sequence of intervals equals a point

  1. Dec 6, 2012 #1
    Intersection of a sequence of intervals equals a point (Analysis)

    1. The problem statement, all variables and given/known data
    Let A[itex]_{n}[/itex] = [a[itex]_{n}[/itex], b[itex]_{n}[/itex]] be a sequence of intervals s.t. A[itex]_{n}[/itex]>A[itex]_{n+1}[/itex] and |b[itex]_{n}[/itex]-a[itex]_{n}[/itex]|[itex]\rightarrow[/itex]0. Then [itex]\cap^{∞}_{n=1}[/itex]A[itex]_{n}[/itex]={p} for some p[itex]\in[/itex]R.
    2. Relevant equations
    Monotonic Convergent Theorem
    If {a[itex]_{n}[/itex]} is a sequence of real numbers, then this sequence has a finite limit if and only if the sequence is bounded.

    R = Real Numbers
    N = Natural Numbers

    3. The attempt at a solution
    Since [a[itex]_{m}[/itex], b[itex]_{m}[/itex]]≤[a[itex]_{n}[/itex],b[itex]_{n}[/itex] for n≤m, then a[itex]_{n}[/itex]≤a[itex]_{m}[/itex] for n≤m and implies {a[itex]_{n}[/itex]}, n[itex]\in[/itex]N is monotonically nondecreasing.

    Since a[itex]_{m}[/itex]≤b[itex]_{m}[/itex] and a[itex]_{n}[/itex]≤b[itex]_{n}[/itex], then a[itex]_{i}[/itex]≤b[itex]_{j}[/itex] for i,j[itex]\in[/itex]N. And {b[itex]_{j}[/itex]} is monotonically nonincreasing and bounded below [itex]\forall[/itex]a[itex]_{i}[/itex],i[itex]\in[/itex]N.

    Then by Monotonic Convergent Theorem, lim[itex]_{n\rightarrow∞}[/itex]{a[itex]_{n}[/itex]} and lim[itex]_{n\rightarrow∞}[/itex]{b[itex]_{n}[/itex]} exist. Also, lim[itex]_{n\rightarrow∞}[/itex]{a[itex]_{n}[/itex]}=sup(a[itex]_{n}[/itex]}) = α and lim[itex]_{n\rightarrow∞}[/itex]{b[itex]_{n}[/itex]}=inf(b[itex]_{n}[/itex]}) = β.

    Then since α=sup(a[itex]_{n}[/itex]), α≤b[itex]_{n}[/itex], n[itex]\in[/itex]N, β=inf(b[itex]_{n}[/itex]), α≤β, this implies that [α,β]≠0. Since a[itex]_{n}[/itex]≤α≤β≤b[itex]_{n}[/itex], then [α,β][itex]\subset[/itex][a[itex]_{n}[/itex], b[itex]_{n}[/itex]]. This implies [itex]\cap^{∞}_{n=1}[/itex]=[α,β].

    Then lim[itex]_{n\rightarrow∞}[/itex](b[itex]_{n}[/itex]-a[itex]_{n}[/itex])=0. This implies lim[itex]_{n\rightarrow∞}[/itex](b[itex]_{n}[/itex])-lim[itex]_{n\rightarrow∞}[/itex](a[itex]_{n}[/itex])=0. This implies β-α=0, implies β=α.

    Then [α,β] = {α} = {β}, implies [α,β] = {p} for some p[itex]\in[/itex]R.
    ∴ [itex]\cap^{∞}_{n=1}[/itex]A[itex]_{n}[/itex]={p} for some p[itex]\in[/itex]R.
     
    Last edited: Dec 6, 2012
  2. jcsd
  3. Dec 6, 2012 #2

    jbunniii

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    Re: Intersection of a sequence of intervals equals a point (Analysis)

    OK up to here, I think.
    I'm not sure how you concluded that last bit. All this demonstrates is that [itex][\alpha,\beta] \subset \cap_{n=1}^{\infty}[a_n,b_n][/itex].
    I think you have all the necessary ingredients for the proof, but it is more cluttered than it needs to be. It suffices to prove the following two statements:

    1. The intersection is non-empty. For example, it contains [itex]\sup a_n[/itex]. You have already proved this.

    2. The intersection cannot contain more than one point. For suppose it contained two distinct points [itex]x[/itex] and [itex]y[/itex]. Then there is some nonzero distance between them: [itex]|x - y| = d > 0[/itex]. Use the fact that [itex]b_n - a_n \rightarrow 0[/itex] to show that this cannot happen.
     
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