# Prove: $a_n \leqslant x$ $\forall_{n \geqslant 1} \Rightarrow lim_{n \to \infty}$ $a_n \leqslant x$

I'm not sure about how to do the proof of this exercise of my math study. Its exercise 5.4.8 of Analysis I by Terence Tao:

Let $(a_n) _{n=1}^{\infty}$ be a Cauchy sequence of rationals and $x \in \mathbb{R}$

Prove: $a_n \leqslant x$ $\forall_{n \geqslant 1} \Rightarrow lim_{n \to \infty}$ $a_n \leqslant x$

As a hint, they tell me to use a contradiction and the theorem that $\forall$ $x < y \in \mathbb{R}$, $\exists$ $q \in \mathbb{Q}$ for which $x < q < y$

This is what I've done so far:

Let $a_n \leqslant x$ $\forall_{n \geqslant 1}$, and assume $lim_{n \to \infty}$ $a_n \nleqslant x$
$\Rightarrow lim_{n \to \infty}$ $a_n > x$
$\Rightarrow \exists$ $q \in \mathbb{Q}$ with $x < q < lim_{n \to \infty}$ $a_n$

Can you explain me how to complete the prove?

First of all it seems like you're already assuming that the limit exists. Now if this is true, and we assume $\lim_{n\to\infty} a_n=a>x$, then choose $\epsilon =a-x$. Then we find an $N\in\mathbb{N}$ such that for all $n>N$ we have $|a_n-a|<a-x$, hence $a-a_n<a-x$ and we find (for all $n>\mathbb{N}$) $a_n>x$, which is a contradiction.
• $|a_n-a|=\max(a_n-a,a-a_n)$, so $a-a_n\leq |a_n-a|$. If now $|a_n-a|<a-x$ it follows that also $a-a_n<a-x$. – Uncountable Feb 26 '15 at 20:02
Let $u = \lim a_n$ and suppose $u > x$. Given $\epsilon := u - x$, there exists an $k \in \Bbb N$ such that $a_k > u - \epsilon$, i.e., $a_k > x$. This contradicts the assumption that $a_n \le x$ for all $n \in \Bbb N$.
• Hi @Peter, since $\lim_{n\to \infty} a_n = u$, for every $\epsilon > 0$, there exists a $k \in \Bbb N$ such that $|a_n - a| < \epsilon$ for all $n \ge k$, i.e., $u - \epsilon < a_n < u + \epsilon$ for all $n \ge k$. So choosing $\epsilon = u - x > 0$, we know that there is a $k \in \Bbb N$ such that $u - \epsilon < a_k < u + \epsilon$. In particular, $a_k > u - \epsilon$. Since $\epsilon = u - x$, $u - \epsilon = x$. Therefore $a_k > x$. – kobe Feb 26 '15 at 19:51