Exercise 2.2.1

Question

What happens if we reverse the order of the quantifiers in Definition 2.2.3?

Definition: A sequence $\left(x_{n}\right)$ \emph{verconges} to $x$ if \emph{there exists} an $\epsilon>0$ such that \emph{for all} $N \in \mathbf{N}$ it is true that $n \geq N$ implies $\left|x_{n}-x\right|<\epsilon$

Give an example of a vercongent sequence. Is there an example of a vercongent sequence that is divergent? Can a sequence verconge to two different values? What exactly is being described in this strange definition?

Ulisse Mini · Accepted Answer

Firstly, since we have \emph{for all} $N \in \mathbf{N}$ we can remove $N$ entirely and just say $n \in \mathbf{N}$. Our new definition is

Definition: A sequence $(x_n)$ \emph{verconges} to $x$ if \emph{there exists} an $\epsilon > 0$ such that $\emph{for all}$ $n \in \mathbf{N}$ we have $|x_n - x| < \epsilon$.

In other words, a series $(x_n)$ \emph{verconges} to $x$ if $|x_n - x|$ is bounded. This is a silly definition though since if $|x_n - x|$ is bounded, then $|x_n - x'|$ is bounded for all $x' \in \mathbf{R}$, meaning if a sequence is vercongent it verconges to every $x' \in \mathbf{R}$.

Put another way, a sequence is vercongent \emph{if and only if} it is bounded.

Exercise 2.2.1

Answers

Comments

Exercise 2.2.1

Answers

Comments

Add answer