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?

Thanks in advance!


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.

  • $\begingroup$ Thank you for your answer. It's almost clear to me. The only thing I don't understand is why a − an < a − x follows from |an − a| < a − x $\endgroup$ – Peter Feb 26 '15 at 19:48
  • $\begingroup$ $|a_n-a|=\max(a_n-a,a-a_n)$, so $a-a_n\leq |a_n-a|$. If now $|a_n-a|<a-x$ it follows that also $a-a_n<a-x$. $\endgroup$ – 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 $n \in \Bbb N$.

  • $\begingroup$ I understand what you mean. But why can we say that there exists such a k? $\endgroup$ – Peter Feb 26 '15 at 19:41
  • $\begingroup$ 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 $a_k > x$. $\endgroup$ – kobe Feb 26 '15 at 19:51

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