seemingly simple limit proof

**DontKnoMaff** · October 2nd 2010, 09:00 AM

Problem Statement:
suppose that { $s_n$ } and { $t_n$ } are sequences such that $s_n \leq t_n$ for all $n$ . Show that if $s_n \rightarrow s$ and $t_n \rightarrow t$ , then $s \leq t$ . Hint: an " $\varepsilon$ -argument" works well here.

Relevant Theorems:
Thm 3.3-a (Baby-Rudin) - $\lim_{n \rightarrow \inf}(s_n+t_n)=s+t$
My Work:
For some $N_1$ , when $n \geq N_1$ , then $|s-s_n| < \varepsilon$ given $\varepsilon > 0$ .
For some $N_2$ , when $n \geq N_2$ , then $|t- t_n| < \varepsilon$ given $\varepsilon > 0$ .
now, $\lim_{n \rightarrow \inf}s_n=s$ and $\lim_{n \rightarrow \inf}t_n=t$ .
From thm 3.3, we have that $\lim_{n \rightarrow \inf}(t_n-s_n)=t-s$ and since $(t_n-s_n) \geq 0$ for all $n$ , then $t-s \geq 0$

**Bruno J.** · October 2nd 2010, 10:34 AM

Originally Posted by DontKnoMaff

Problem Statement:
suppose that { $s_n$ } and { $t_n$ } are sequences such that $s_n \leq t_n$ for all $n$ . Show that if $s_n \rightarrow s$ and $t_n \rightarrow t$ , then $s \leq t$ . Hint: an " $\varepsilon$ -argument" works well here.

Relevant Theorems:
Thm 3.3-a (Baby-Rudin) - $\lim_{n \rightarrow \inf}(s_n+t_n)=s+t$
My Work:
For some $N_1$ , when $n \geq N_1$ , then $|s-s_n| < \varepsilon$ given $\varepsilon > 0$ .
For some $N_2$ , when $n \geq N_2$ , then $|t- t_n| < \varepsilon$ given $\varepsilon > 0$ .
now, $\lim_{n \rightarrow \inf}s_n=s$ and $\lim_{n \rightarrow \inf}t_n=t$ .
(*) From thm 3.3, we have that $\lim_{n \rightarrow \inf}(t_n-s_n)=t-s$ and since $(t_n-s_n) \geq 0$ for all $n$ , then $t-s \geq 0$

How does (*) follow from any of the above? What you're saying is true (that if a convergent sequence is always non-negative, then its limit is non-negative), but it's equivalent to what you're trying to show. And I don't see how it follows from what you've written. (What have you done with $\epsilon, N_1, \mbox{and }N_2$ , apart from summoning them into existence?)

A proof by contradiction works well. Suppose that $a_n \geq 0 \ \forall n$ and that $a_n \to a$ . Suppose that $a<0$ . Take $\epsilon>0$ small enough that $a+\epsilon < 0$ . What do you get if you find $N$ large enough that $|a_n-a|<\epsilon$ for $n\geq N$ ? (Break up the absolute value into two inequalities; one of them will contradict the non-negativity of $a_n$ ).

Then take $a_n$ to be the sequence $t_n-s_n$ , as you did.

**DontKnoMaff** · October 2nd 2010, 11:58 AM

thanks alot. I am horrible at writing these proofs. OK here is what I've got now:

take { $a_n$ } where $a_n=t_n-s_n$ for all $n$ and { $a_n$ } $\rightarrow a=t-s$ .
We have assumed that $s_n \leq t_n$ which yields $t_n-s_n \geq 0$ , so $a_n \geq 0$ for all $n$ .
We now want $a>0$ .
Suppose $a<0$ , take $\epsilon > 0$ s.t. $a+\epsilon < 0$ .
Now take $N$ s.t. when $n \geq N$ , then , $|a_n-a| < \epsilon$ .
Now since $|a| > \epsilon$ and $a <0$ and $a_n >0$ , this is a contradiction and so $a \geq 0$ which gives $t-s \geq 0$

I think that works now.

**Bruno J.** · October 2nd 2010, 12:09 PM

Everything is fine until the last line.
What you should have written is that since $|a_n-a| < \epsilon$ , we have $-\epsilon < a_n-a < \epsilon$ (this is by definition of the absolute value). But $a_n-a < \epsilon \Rightarrow a_n < a+\epsilon < 0$ , which contradicts $a_n \geq 0$ .

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