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I have two sequences $$1,2,3,1,2,3,1,2,3,1,2,3,\cdots$$ and the second sequence is $(-1)^n$.

I am saying that both the sequence have no limit point but my friend is saying both the sequence have limit point

Which is correct, please help...

Nosrati
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Abhishek Chandra
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3 Answers3

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The definition is: A point $l$ is said to be a limit point of a sequence $\{a_n\}$ if every neighborhood of $l$ contains infinite terms of the sequence. The limit points of first sequence are $1,2,3$ and for the 2nd sequence the limit points are $-1$ and $1$. As every neighborhood of $1$ contains infinite terms: namely $a_{3k+1},~ k=0,1,2...$ so $1$ is a limit point of the first sequence. Similarly you can do for other points

Arun
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  • Really? Writing nbd because neighborhood is too much work? – 5xum Feb 01 '17 at 09:55
  • @5xum Is the explanation wrong?? – Arun Feb 01 '17 at 09:57
  • Please elaborate..I think something is missing – Abhishek Chandra Feb 01 '17 at 09:59
  • @AbhishekChandra I think you have asked about limit points of the sequence and not limit of the sequence. Clearly these two sequences have no limits but aren't $1,2,3$ the limit points of the first sequence?? If not, then what is the definition of limit point of a sequence?? – Arun Feb 01 '17 at 10:02
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Terminology: "Limit" and "limit point" are different things. (The style in German would be to make "limitpoint" a new word, to distinguish it from "limit".)

A sequence $(a_n)_n$ converges to the limit $a$ iff for all $r>0$ the set $\{n:|a-a_n|\geq r\}$ is finite. Equivalently that for all $r>0$ the set $\{n: |a-a_n|<r\}$ is co-finite in $\mathbb N.$ (A set $S\subset \mathbb N$ is co-finite in $\mathbb N$ iff $\mathbb N$ \ $S$ is finite.)

A value $b$ is a limit point of $(a_n)_n$ iff for all $r>0$ the set $\{n:|b-a_n|<r\}$ is infinite.

A sequence may have many limit points, or none. A sequence $(a_n)_n$ has exactly one limit point $a$ iff $(a_n)_n$ converges to $a$

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Neither sequence has a limit because for both sequences, the claim

$$\exists L: \forall \epsilon > 0 \exists N\in\mathbb N\forall n\in\mathbb N: n>N\implies |L-a_n|<\epsilon$$

or, in writing:

There exists an $L$ such that, for all $\epsilon>0$, there exists some $N$ such that $|L-a_n|<\epsilon$ is true for all $n>N$

is false.


You can show the claim is false by showing that its negation,

$$\forall L\exists \epsilon > 0\forall N\in \mathbb N\exists n>N: |L-a_n|\geq\epsilon$$

is true.


An easier way to disprove the claim, of course, is to show that both sequence have two subsequences with different limits, and that is only possible if the sequence is not convergent.

5xum
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