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Let $(x_n)$ and $(y_n)$ be sequences of real numbers that satisfy $\lim x_n =a$ and $\lim y_n=b$.

Prove that if $a<b$ then $\exists n_0 \in \mathbb{N}$ such that $\forall n > n_0$, $x_n < y_n$

I found this exercise at Elon Lages Lima's book Real Analysis vol. 1, chapter 3, and wrote the following proof, I would like to know if it is correct:

First of all, we already know from the definition of limit that \begin{align} \forall \epsilon >0, \exists n_1 \in \mathbb{N}\; &\text{s.t.}\; \forall n>n_1, |x_n-a|<\epsilon\\ \forall \epsilon >0, \exists n_2 \in \mathbb{N}\; &\text{s.t.}\; \forall n>n_2, |y_n-b|<\epsilon \end{align}

Now let's suppose that $x_n>y_n$ for sufficiently big $n$ and try to prove it by absurd. We get the following inequality for $\forall n > \max\{n_1,n_2\}:$ \begin{align} b-\epsilon < y_n < x_n < a + \epsilon \end{align} Choosing $\epsilon = a-b$, we get: \begin{align} b-(a-b)<y_n&<x_n<a+(a-b)\\ 2b-a &< 2a-b\\ 3b &< 3a \\ b &<a \end{align} which contradicts the first premise that $a<b$, thus proving our proposition.

Thanks in advance.

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    $\begingroup$ Why did you choose $\epsilon=a-b$ since $a<b$. $\endgroup$ Commented Jul 20, 2020 at 23:17
  • $\begingroup$ Indeed, I did not noticed this mistake. I have to be more careful to not choose negative epsilons again. Your answer really helped me. $\endgroup$ Commented Jul 20, 2020 at 23:21

2 Answers 2

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Direct approach

Assume that $a<b$.

With $$\epsilon=\frac{b-a}{2}$$ we can say $$(\exists N_1\in \Bbb N) \; : \;(\forall n\ge N_1)$$ $$ x_n<a+\frac{b-a}{2}$$ and

$$(\exists N_2\in \Bbb N)\;:\;(\forall n\ge N_2)\;$$ $$ b-\frac{b-a}{2}<y_n$$

then with $ N=\max(N_1,N_2)$, we have

$$(\forall n\ge N)\; \; x_n<\frac{a+b}{2}<y_n$$

Done.

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I would try this in a different way. Since $b>a$, than $b-a=r>0$. Then, from the definition, there exists $n_0$ and $n_1$ such that $|x_n-a|<r/2=(b-a)/2$ for all $n\geq n_0$ and $|y_n-b|<(b-a)/2$ for all $n\geq n_1$. Then, for $n\geq \max\{n_0,n_1\}$ the following is true $x_n<(b-a)/2+a=(b+a)/2$ and $y_n>-(b-a)/2+b=(b+a)/2$. Finally, $x_n<(b+a)/2<y_n$ for all $n\geq \max\{n_0,n_1\}$.

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