A negation of convergence statement for subsequences

78 Views Asked by At

Given the following definition: A number $c$ is a ${\bf limit \; point}$ of the sequence $\{ a_n \}$ if there exists a subsequence $\{ a_{n_k} \}$ such that $\lim_{k \to \infty} a_{n_k} = c$. Let $S$ be the set of all limit points of $(a_n)$.

Im trying to understand precisely what $x \notin S$ means in terms of quantifiers.

thought:

It means that $\exists \epsilon > 0$ so that $\forall N$ we can always find $n > N$ so that $|a_{n_k} - x | \geq \epsilon $ and in particular $|a_n - x | \geq \epsilon$ because $a_{n_k}$ is any subsequence of $a_n$

Is my thought process correct?

1

There are 1 best solutions below

4
On BEST ANSWER

You are given $(a_n)$ and told that $x$ is not a limit point. No subsequence $(a_{n_k})$ is given to you. So your interpretation is not correct.

The correct interpretations is there exists $\epsilon >0$ and a positive integer $m$ such that $|a_n -x| >\epsilon$ for all $n>m$. [Existence of a subsequence converging to $x$ is equivalent to the fact that for every $\epsilon >0$ the inequality $|a_n-x| <\epsilon$ holds for infinitely many values of $n$].