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In general, when there are multiple objects around which a sequence, function, or set accumulates, the inferior and superior limits extract the smallest and largest of them.

This is quoted from wikipedia. I know limsup to be limit of the upper bound of a sequence. But I am unable to associate the definition of limsup to that of accumulation point. So, can anyone tell me how are they related to each other? And also why is limsup to be the largest accumulation point?? Can anyone also tell what the wiki statement wants to tell?

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    $\begingroup$ Can you prove that the limsup (if finite) is an accumulation point? Can you prove that no accumulation point can be larger than the limsup? $\endgroup$ Jun 1 '15 at 13:18
  • $\begingroup$ @Henning Makholm: Hi sir, unfortunately, I don't know. Actually I am really confused what accumulation point is. . $\endgroup$
    – user142971
    Jun 1 '15 at 13:20
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Let's deal with a sequence for simplicity. The set of accumulation points of a sequence $\{ x_k \}$ is the set of limits of convergent subsequences of $\{ x_k \}$. I claim that the limsup is one of these limits. Let's prove it.

By definition, $\limsup_{k \to \infty} x_k = \lim_{n \to \infty} \sup_{k \geq n} x_k$. For each $n \in \mathbb{N}$, use the definition of supremum to find $k_n \geq n$ such that $x_{k_n} \geq \left ( \sup_{j \geq n} x_j \right ) - 1/n$. Then $x_{k_n}$ converges to the limsup, because

$$\left ( \sup_{j \geq n} x_j \right ) - 1/n \leq x_{k_n} \leq \left ( \sup_{j \geq n} x_j \right )$$

so by the squeeze theorem

$$\limsup_{n \to \infty} x_n \leq x_{k_n} \leq \limsup_{n \to \infty} x_n.$$

You said you understand that the limsup is an upper bound on the accumulation points, so I won't prove that; let me know if I misunderstood you.

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  • $\begingroup$ Sir, can you make me a favour? I have got two definitions of accumulation points: 1)a point that has at least one member of the sequence in its neighbourhood 2)a point which contains infinitely many members of the sequence in its neighbourhood. So, what is it actually? What does this mean when one says that the sequence is accumulating around a point?I'll be grateful if you help me clarify the meaning of accumulation points. $\endgroup$
    – user142971
    Jun 1 '15 at 13:38
  • $\begingroup$ And BTW, sir, I didn't say that limsup is an upper bound on the accumulation points; what I said that I know what limsup is. $$\lim\sup\{x_n\} = \lim_{n \to \infty} \inf_{n} \{x_n\}$$. I am not aware of the statement you wrote. $\endgroup$
    – user142971
    Jun 1 '15 at 13:50
  • $\begingroup$ @user36790 Neither of your definitions are correct as stated. The second one is close. A correct statement is that $x$ is a limit point of $\{ x_n \}$ if every neighborhood of $x$ contains infinitely many members of $\{x_n\} $. This is equivalent to the sequential version that I said. If you're still confused, I'd suggest looking at some of the classic examples: any convergent sequence, $(-1)^n$, and an enumeration of $\mathbb{Q} \cap [0,1]$ are good places to start. $\endgroup$
    – Ian
    Jun 1 '15 at 14:16
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    $\begingroup$ @user36790 You should try and read some things about subsequences in your analysis text. It is too large a subject for an MSE question like this. $\endgroup$
    – Ian
    Jun 1 '15 at 17:04
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    $\begingroup$ @user36790 Strichartz The Way of Analysis was my undergrad text. There are many such undergrad analysis texts. $\endgroup$
    – Ian
    Jun 1 '15 at 17:47

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