Find the Set of All Accumulation Points (also called Limit Points) for the given set $S$.
$S = \{a_n:n\in \Bbb{N}\}$ where $a_n = \left\{ \begin{array}{c} 0\text{, if } n \text{ is odd} \\ \dfrac{n}{n+1}\text{, if } n \text{ is even} \end{array} \right..$
In my perspective there are two possible answers.
1) $L(S) = \{0,1\}$
Using the definition of the accumulation or limit point, of a sequence. The number $a$ is said to be an accumulation point of $(a_n)$ if there exists a subsequence such that the limit of the subsequence is equal to $a$.
2) $L(S) = \{1\}$
Using the definition of the accumulation or limit point of a set.
The point $a$ is an accumulation point of the set $S$ if each neighborhood of $a$ contains a point of $S$, other than $a$.
$1$ is the only accumulation point because not all the neighborhoods of $0$ contain a point of the set $S$ other than $0$.
As the question mentions "the given set $S$", I vote for the second answer.
What do you think?
Best Answer
The critical question is whether $S$ is a set or a sequence. You call it a set, but defining values for $a_n$ for all $n$ makes it sound like a sequence. If it is a set, it can have only one member $0$. If it is a sequence, all the odd positions are $0$. As a sequence, it has an accumulation point at $0$ and an accumulation point at $1$ because you can select a subsequence converging to each of those two points. It does not have a limit because there are members of the sequence with arbitrarily high indices that are far from $0$ and others that are far from $1$. As a set, the only accumulation point is $1$. Sets are not considered to have limits because they lack the concept of order.