# Thread: seemingly simple limit proof

1. ## seemingly simple limit proof

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 $t-s \geq 0$

2. 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.

3. 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.

4. 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 $a_n \geq 0$.