1

Let $\left\{ x_{n}\right\} _{n}$ be a sequence. Define

$E_0=${$r\in\mathbb{R}$:$\lim_{k\rightarrow\infty}$ $x_{n_{k}}=r$}.

If $\left\{ x_{n}\right\} _{n}$ has a subsequence converging to $\pm \infty$ we add this to $E_0$ to obtain $E$. So,

$E=${$x\in\mathbb{R}\cup${$\pm \infty$}:$\lim _{k\rightarrow \infty }x_{n_{k}}=x$}.

Remark. If $\left\{ x_{n_{k}}\right\} _{k}$ is increasing then sup$_{k} $$x_{n_{k}}\in E$, if not inf$_{k}$ $x_{n_{k}}\in E$.

I couldn't understand this remark that can you explain more clearly?

Chill2Macht
  • 22,055
  • 10
  • 67
  • 178
  • It is simply noting that if $x_{n_k}$ increases then $$\lim_{k\to\infty} x_{n_k}=\sup_k x_{n_k}$$ and similarly the other parts describe the cases when $x_{n_k}$ decreases. – String Mar 19 '17 at 23:24
  • I don't think it's right - the sequence $0, -1, -1/2, -1/3, \ldots$ is not increasing and it has inf $-1$ but no subsequence converges to $-1$. – Daniel Schepler Mar 19 '17 at 23:25
  • @DanielSchepler: Ah, you are absolutely correct. I misread that part. – String Mar 19 '17 at 23:30

1 Answers1

2

So $\{x_{n_k}\}_k$ is increasing. If this series is bounded, we know it will converge to a number $M$, otherwise, it will go to infinity. Either way, we could say $\sup_k\{x_{n_k}\}=\lim_{k\rightarrow \infty}=x,\,x\in \mathbb R \cup\{-\infty,+\infty\}$.

Similar logic could be applied to decreasing and $\inf$.

However, I do not think the "if not" in the remark would be correct - not increasing does not mean decreasing. If it is oscillating, $\inf$ always exists in $\mathbb R \cup\{-\infty,+\infty\}$, but $\lim$ does not necessarily exist.

Chill2Macht
  • 22,055
  • 10
  • 67
  • 178
Jay Zha
  • 7,992