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Given bounded sequence $\{a_n\}_{n=1}^{\infty}$ which doesn't have minimum or maximum - prove that $a_n$ doesn't converge.

Thanks a lot.

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Something is wrong with the statement: $a_n=1/n$ is bounded, doesn't have minimum and maximum value in its sequence, but it does converge. –  lhf May 2 '12 at 11:39
    
What is this $L$? –  Joe Johnson 126 May 2 '12 at 11:40
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@lhf: It has a maximum of $1$. –  Joe Johnson 126 May 2 '12 at 11:40
    
@lhf: I think the OP means minimum or maximum. –  Joe Johnson 126 May 2 '12 at 11:41
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I understand the assumptions as follows: If $\{a_n\}_n$ converges, then either $\inf_n a_n$ or $\sup_n a_n$ is attained. –  Siminore May 2 '12 at 11:48
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Hint: $a_n$ is bounded and therefore, it has an infimum and a supremum. Since it does not have a maximum and minimum, it has infinitely many elements near both infimum and supremum. This means that there are 2 subsequences of $a_n$; one of them converges to the infimum and the other converges to the supremum. But infimum and supremum are distinct (if they were not distinct it would be a constant sequence). So, $a_n$ does not converge.

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I understand the idea behind the proof; how can I show for example that there is a subsequence of $a_n$ that converges to the supremum? –  Anonymous May 2 '12 at 12:04
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The supremum is a cluster point of $\{a_n\}_n$, so there exists a subsequence converging to it. –  Siminore May 2 '12 at 12:08
    
what's the difference between cluster point and an accumulation point? –  Anonymous May 2 '12 at 12:21
    
I just mean that the supremum and the infimum of a subset $E \subset \mathbb{R}$ belong to the closure $\overline{E}$. –  Siminore May 2 '12 at 12:57
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Assume that $a_n \to 0$, just for simplicity. Call $M=\sup_n a_n$ and $m = \inf_n a_n$, so that $M$ and $m$ are finite by assumption, and $m \leq 0 \leq M$. If $M=m$, there is nothing to prove. Otherwise, fix $\epsilon>0$ small, and assume that $m \leq 0 < M$. There exists $N \in \mathbb{N}$ such that $-\epsilon < a_n < \epsilon$ whenever $n > N$. Therefore either $m = \inf \{a_n \mid n \leq N\}$ or $M=\sup \{a_n \mid n \leq N\}$, since $a_n$ can't be arbitrarily close to a number $m \leq 0$ and to a number $M>0$. Hence either $M$ or $m$ is attained.

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are you proving this by contradiction? if so, what do you contradict? –  Anonymous May 2 '12 at 12:16
    
I proved that any converging sequence attains either its infimum or its maximum. –  Siminore May 2 '12 at 12:56
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