prove the limit inferior of $(x_n)$ where $n \in\mathbb{N}$

Question

The problem states let $(x_n)$ be a bounded sequence for each $n \in\mathbb{N}$. Let $t_n=inf\{x_k: k\geq n\}$. Prove that $(t_n)$ is monotone and convergent.

After a little research because I was confused what limit inferior really was this was the proof I tried coming up with. Proof: Since $(x_n)$ is bounded then it has a supremum and info um. Let $t_n=inf\{x_n: k\geq n\}=inf\{x_n,x_{n+1},...\}$. It follows that $t_{n+1}=inf\{x_{n+1}: k\geq n+1\}=inf\{x_{n+1},x_{n+2},...\}$ If we let set $A= \{x_{n+1},...\}$ and set $B=\{x_n,...\}$ (Note* I assume that set A and B have greatest lower bounds since these are subsets of $(x_n)$ and $(x_n)$ is bounded). Then it follows that $A \subset B$ and $inf A\geq inf B$ Otherwise if $inf A<inf B$ then $inf A \leq x_n$ a contradiction since $x_n \notin B$. It follows that $x_{n+1} \geq x_n$ otherwise if $x_n> x_{n+1}$ then $inf A \leq x_n$ a contradiction from above. Hence $t_n$ is increasing and by the Monotone convergence theorem $t_n$ is monotone and convergent. Not sure If I have all this correct.

«Since $(x_n)$ is bounded then it is monotone». That is most certainly not true. Just consider the sequence with $x_n=(-1)^n$ for all $n$. (You should probablt review the statement of the Monotone convergence theorem, because it does not say anything even close to that!) — Mariano Suárez-Álvarez, Jul 05 '13 at 06:08

score 2 · Accepted Answer · answered Jul 05 '13 at 06:11

2

I suggest you prove

First: if $A$ and $B$ are bounded subsets of $\mathbb R$ and $A\subseteq B$, then $\inf A\geq\inf B$.

and then rethink a bit your argument, trying to make it a little more organized.

answered Jul 05 '13 at 06:11

Mariano Suárez-Álvarez

135,076

Oh OK. I thought what I had was correct. – user60887 Jul 05 '13 at 14:10
To prove that statement I can suppose that $inf A < inf B$ and assume that $A \subseteq B$ which follows that every $x\in A$ is an element in Set B. Then $\inf\ B \leq x$. which follows that $\inf\ B \leq \inf\ A$ a contradiction. – user60887 Jul 05 '13 at 15:19
@user60887, you can do this directly, using the definition of the $\inf$ —there is no need to use an argument by contradiction. – Mariano Suárez-Álvarez Jul 05 '13 at 17:40
I thought it be easier by contradiction. I don't think it matters what proof technique you use. – user60887 Jul 05 '13 at 19:24
Feel free to use as complicated an argument as you wish! – Mariano Suárez-Álvarez Jul 05 '13 at 19:25
but would my proof work though? – user60887 Jul 05 '13 at 19:27
The argument in your comment above does not seem to prove anything: the second sentence does not seem to do anything, really (what is $x$ in that sentence?) – Mariano Suárez-Álvarez Jul 05 '13 at 19:29
Oops $x$ is supposed to be an arbitrary element in Set $A$ – user60887 Jul 05 '13 at 19:35
Moreover, you never used the hypothesis that $\inf A<\inf B$. – Mariano Suárez-Álvarez Jul 05 '13 at 19:36
I need to elaborate on my second sentence. Since every element of A is an element of set B it follows that the $inf B\leq x$. Thus $inf B$ is a lower bound for set $A$ which means $inf B \leq inf A$ which is a contradiction. – user60887 Jul 05 '13 at 19:58
Again, you never used your hypothesis thatr $\inf A<\inf B$ for anything. – Mariano Suárez-Álvarez Jul 05 '13 at 20:00
I think i would be best off using what I said in a direct proof. – user60887 Jul 05 '13 at 20:01
1

Indeed :-) (My point all along was that,actually, your proof is a direct proof) – Mariano Suárez-Álvarez Jul 05 '13 at 20:03
Thanks alot for you help. I think from here I can fix up my proof above. – user60887 Jul 05 '13 at 20:05

score 1 · Answer 2 · answered Jul 05 '13 at 06:30

It is well known that an increasing sequence bounded above converges.

Since $(x_n)$ is bounded above $(t_n)$, always being a lower bound of some subset of $\{x_n : n \in \mathbb{N}\}$, is also bounded above.

Let $A_n = \{x_k : k \ge n\}$. We have $A_{n+1} \subseteq A_n$, and so for any $x \in A_{n+1}$, $t_n \le x$. Therefore, $t_n$ is a lower bound of $A_{n+1}$, so by definition,

$$t_n \le \inf A_{n+1} = t_{n+1}$$

prove the limit inferior of $(x_n)$ where $n \in\mathbb{N}$

2 Answers2