# Inequality involving two convergent sequences

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 $$a then $$\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) which contradicts the first premise that $$a, thus proving our proposition.

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

Direct approach

Assume that $$a.

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

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

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

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

Done.

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 $$|x_n-a| for all $$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 for all $$n\geq \max\{n_0,n_1\}$$.