# Proof for Convergent Sequences [duplicate]

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Did I underestimate the limit proof?

Let $(a)_{n\in \Bbb N}$ and $(b)_{n\in \Bbb N}$ be sequences of real numbers such that $a_n$ $\le$ $b_n$ for all $n\in \Bbb N$. Prove that if $a_n \to a$ and $b_n \to b$; then a $\le$ b.

I have this question as homework. I have some sort of solution in mind but the professor wants us to hand in a formal proof. What I thought so far is to assume contrary, that a $\gt$ b, and to examine $n$ for large numbers to conclude $b_n \gt a_n$ and to have contradiction. But my problem is I don't even know what formal proof is. Could you please tell me what it is using this question.

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 Consider the sequence $b_n - a_n$. What does this converge to? – Brian Oct 14 '12 at 18:24 b - a. Where does this lead me to? – Mert Toka Oct 14 '12 at 18:51

## marked as duplicate by Qiaochu YuanOct 15 '12 at 5:01

Assume by contradiction that $a >b$.

Let $\epsilon >0$, we will pick it later.

Then there exists an $N$ so that for all $n> N$ we have

$$|a_n -a| < \epsilon \Rightarrow a_n > a- \epsilon$$

$$|b_n -b| < \epsilon \Rightarrow b_n < b+\epsilon$$

Now, if you pick some $\epsilon$ so that $b+ \epsilon \leq a-\epsilon$ you get

$$b_n < b+ \epsilon \leq a-\epsilon < a_n \,,$$

You figure now the right $\epsilon$, and start your argument by "Let $\epsilon= ...$.

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 The correct $\epsilon$ is $a-b$ to get: $b_n < a \leq b < a_n$ ? – CodeKingPlusPlus Oct 14 '12 at 19:23 I suppose $a - b$ can be used for the conradiction. Is it what you have in mind @n-s ? – Mert Toka Oct 14 '12 at 20:04 @MertToka $b+\epsilon \leq a- \epsilon$ means $2\epsilon \leq a-b$... So nope $\epsilon = a-b$ won't work, but figuring one so that $2 \epsilon \leq a-b$ should be really easy.... – N. S. Oct 14 '12 at 20:44

Proof:

Let $a_n$ and $b_n$ be sequences such that $\forall_n a_n \leq b_n$ and $a_b \to a$ and $b_n \to b$.

Suppose $a \nleq b$, that is $a > b$. We already know:

$\forall_{\epsilon_1 > 0} \exists_N \forall_{n>N_1} \implies |a_n - a| < \epsilon_1$

$\forall_{\epsilon_2 > 0} \exists_N \forall_{n>N_2} \implies |b_n - b| < \epsilon_2$

Define $b^* = \min\{|b_n -b| < \epsilon_2\}$ and $a^* = \max\{|a_n - a| <\epsilon_1\}$.

This guarantees that $a^* \in a_n$ and $b^* \in b_n$ and $b^* < a^*$. But $a_n \leq b_n \forall_n$. $\rightarrow\leftarrow$. (I think this is a valid contradiction? I am unsure whether the max and min correspond directly to the index $n$, but maybe this jogs your brain a little more in terms of formality)

Therefore we have a contradiction and so the supposition is false and the original claim is true.

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