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Prove that if x > limsup s_n, then x is not the limit of any subsequence

  1. Aug 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Directly from the definition, for a sequence [tex](s_n)_{n \in \mathbb{N}} \subseteq \mathbb{R}[/tex] prove that if [tex]x > \limsup s_n [/tex], then [tex]x[/tex] is not the limit of any subsequence of [tex](s_n)[/tex]. (i.e. Do not use the fact that [tex]\limsup s_n[/tex] is the supremum of the set of subsequential limits.)


    2. Relevant equations
    I have been told by my instructor that my proof will fail due to problems with inequalities --- but I fail to see where it would fail; i.e. are there any errors where [tex]>[/tex] should be [tex]\ge[/tex] or vice-versa?
    1. The problem statement, all variables and given/known data

    3. The attempt at a solution
    Let's first show that for any sequence [tex](s_n)[/tex] and any number [tex]M[/tex], if [tex]s_n > M[/tex] for all [tex]n \in \mathbb{N}[/tex], then [tex]\limsup s_n \geq M[/tex]. Since [tex]s_n > M[/tex] for all [tex]n \in \mathbb{N}[/tex], then it follows that [tex]s_n > M[/tex] for all [tex]n > N[/tex] for some [tex]N \in \mathbb{N}[/tex]. Now, let's denote [tex]v_N = \sup\{s_n : n > N\}[/tex]. Since [tex]v_N[/tex] is the least upper bound for [tex]s_n[/tex] when [tex]n > N[/tex], then it follows that we have [tex]v_N \geq s_n > M[/tex] for [tex]n > N[/tex]. And note that we have the property [tex]v_N \geq v_{N+1} \geq \ldots[/tex] and so [tex](v_N)[/tex] is monotonically decreasing and since [tex](s_n)[/tex] is bounded, by the monotone convergence theorem, [tex]\lim_{N \to \infty}v_N[/tex] exists and is real. Moreover, this is precisely the definition of the limit superior and hence, we have that [tex]\lim_{N \to \infty}v_N = \lim_{N \to \infty}\sup\{s_n : n > N\} = \limsup s_n[/tex]. Thus, we have [tex]\limsup s_n > M[/tex] as desired.

    For contradiction, suppose that [tex]x[/tex] is the limit of some subsequence [tex](s_{n_k})[/tex] of the sequence [tex](s_n)[/tex]. Let's first denote [tex]v = \limsup s_n[/tex]. Note that if [tex]\lim_{k \to \infty}s_{n_k} = x > v[/tex], then there exists some [tex]N[/tex] such that for all [tex]k > N[/tex], we have [tex]s_{n_k} > v[/tex]. Now, consider [tex]s_{n_k} > v[/tex] for all [tex]k > N[/tex]. Then by the result above, and since removing finite number of terms [tex]1 \leq k \leq N[/tex] does not affect convergence results, it implies [tex]\limsup s_{n_k} > v = \limsup s_n[/tex] --- contradiction, since [tex]\{s_{n_k}: k > N\} \subseteq \{s_n: n > N\}[tex], it implies [tex] \sup\{s_{n_k} : k > N\} \leq \sup\{s_n : n > N\}[tex] and thus [/tex] \lim_{N \to \infty}\sup\{s_{n_k} : k > N\} \leq \lim_{N \to \infty}\sup\{s_n : n > N\}[/tex].
     
  2. jcsd
  3. Aug 12, 2009 #2
    Please ignore this.
     
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