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($a_n$) is a real sequence bounded from above. Let $A :=$ {$s \in \Bbb R:$ $s$ is the limit of a convergent subsequence of ($a_n$) }.

Suppose the supremum of A is a real number, prove by contradiction that the limit superior of ($a_n$) is also a real number, i.e. show that $ \limsup _{n \to \infty} ~ a_n$ is also a real number.

Knowing that $Sup A = a \in \Bbb R$, I need to prove that $$ \limsup _{n \to \infty} ~ a_n$$ is equal to some real number $b$. My understanding of prove by contradiction means: suppose $\lim sup_{x \to \infty} a_n $ is not a real number, then the assumption won't hold. Is this right, and I'm not sure about the following steps. Could someone provide a proof please? Thanks.

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  • $\begingroup$ @TimRaczkowski I've edited the question, feel free to improve the phrasing. $\endgroup$
    – user78290
    Jan 28 '16 at 4:01
  • $\begingroup$ I think it's also necessary to tell us what your definition of the Limit Superior is. For an example $\sup A$ is the definition I use. $\endgroup$
    – Ishfaaq
    Jan 28 '16 at 4:16
  • $\begingroup$ It seems that the question should be to show that $\limsup$ exists. If it exists, it has to be a real number because $\limsup_{n\to\infty}a_n=\lim_{n\to\infty}\sup\{a_k\}_{n=k}^\infty$. In other words, $\limsup$ is the limit of a sequence of real numbers. $\endgroup$ Jan 28 '16 at 4:21
  • $\begingroup$ I was given the definition of the limit superior as: $$\limsup a_n = \inf_{\forall m} \sup_{n \ge m} a_n$$ $\endgroup$
    – user78290
    Jan 28 '16 at 22:12
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Since $a_n$ is bounded above we can think of $A$ as the set of subsequential limits of $(a_n)$ in the extended real sense - i.e., we allow $\pm \infty$. Then a standard result in undergrad. analysis (*) is that $\limsup a_n \in A$. Thus it is real. Note that you don't even need $\sup A < \infty$, this is implicit in $(a_n)$ bounded.

* see Rudin's PoMA for a proof (Theorem 3.17 page 56 in the second edition)

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  • $\begingroup$ Thanks. But I think Rudin's proof already assumed $\limsup a_n = Sup A$, which is not assumed in my question. Should I prove this equality first in order to prove $\limsup a_n$ is real? $\endgroup$
    – user78290
    Jan 28 '16 at 19:41
  • $\begingroup$ This is actually his definition of $\limsup$ - again, with $A$ in the extended reals -, the alternate definition is not used in PoMA (at least in Ch1-6). Yes, the easiest way is probably to prove equivalence with Rudin's definition and then use my answer. $\endgroup$
    – B. Freitas
    Jan 28 '16 at 23:24

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