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...
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...
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
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 $a$
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.