Let $(x_n)$ be a bounded sequence and for each $n ∈ ℕ$ let $s_n:=\sup\{x_k \mid k\geq n\}$ and $S:=\inf\{s_n\}.$ Show that there exist a subsequence of $(x_n)$ that converges to $S$.

I have that $(s_n)$ is a bounded decreasing sequence (by intuition, idk how to prove that). Because $(s_n)$ is bounded an monotone, converges to $S$ then exists $N_1 ∈ ℕ$ such that $s_n-S=|s_n-S|<1/2 ∀ n\geq N_1$ In particular $s_{N_1}=\sup\{x_k \mid k\geq N_1\} < S+1/2$ then by definition of supremum exists $n_1\ge N_1$ such that $s_{N_1}>x_{n_1}>s_{N_1} - 1/2$ hence $|x_{n_1} - S|<1$. if I continue the process I get $n_{k+1}>n_k$ such that $|n_{k+1} - S|<1/(k+1)$ then $n_{k+1}\to S$ by squeeze theorem.

First I need help to prove that $s_n$ is bounded an decreasing and if the rest of my proof is correct or I'm missing anything.


\begin{align} s_1 & = \sup \{x_1, x_2, x_3, x_4, x_5, \ldots\} \\ s_2 & = \sup \{x_2, x_3, x_4, x_5, x_6, \ldots\} \end{align} If a number is bigger than all of $x_1,x_2,x_3,x_4,\ldots,$ then it is bigger than all of $x_2,x_3,x_4,\ldots$

In other words, every upper bound of $\{x_1,x_2,x_3,x_4,\ldots\}$ is an upper bound of $\{x_2,x_3,x_4,\ldots\}.$

Therefore $\sup\{x_1,x_2,x_3,x_4,\ldots\}$ is an upper bound of $\{ x_2, x_3, x_4, \ldots\}.$

Therefore $\sup\{x_1,x_2,x_3,x_4,\ldots\} \ge \sup\{x_2,x_3,x_4,\ldots\}.$

And by the same argument, show that $$\text{whenever }n\ge m, \text{ then } \sup\{x_m,x_{m+1},x_{m+2},\ldots\} \ge \sup\{x_n,x_{n+1},x_{n+2},\ldots\}.$$

Hence $(s_n)_{n=1}^\infty$ is a decreasing sequence.

For every $n\in\mathbb N,$ $s_n = \sup\{x_n,x_{n+1},x_{n+2},\ldots\} \ge x_n \ge \inf \{x_1,x_2,x_3,\ldots\};$ therefore the sequence $(s_n)_{n=1}^\infty$ is bounded below by that infimum $S$.

Next, we know that for every $N\in\mathbb k$, the number $S+ 1/k$ is not a lower bound of $\{x_1,x_2,x_3,\ldots\}.$ Thus for some $n_k$ we have $x_{n_k}<S+1/k.$

  • $\begingroup$ Thank you!! and is the rest of the proof correct?? $\endgroup$ – Karen Mar 26 '17 at 21:01
  • $\begingroup$ I think the rest of it look alright. $\endgroup$ – Michael Hardy Mar 26 '17 at 21:04

Sn >=Sn+1 hence Sn is contracting sequence. Sn has a bound As. Inf(Xn)<=Sn<=Sup(Xn) now inf of Sn is S. So for €>0 there exist Sn such that S <=Sn < S+€ for n >= k. So S-€ < S <= Sn < S+€ Hence |S-Sn| < € for n>=k Where one may assume Sn as a subsequence of the given sequence

  • $\begingroup$ Please have a look at how to format mathematics on Math SE if you need to do so. Furthermore, I suggest that you bookmark this very useful MathJax link as a quick reference for future posts. By the way, Welcome to Math SE, Cheers! $\endgroup$ – user409521 Mar 27 '17 at 2:14

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