Terence Tao Exercise 5.4.8: Boundedness of Limit.

Question

Let $ \{ a_{n} \}_{n=1}^{\infty} $ be a Cauchy sequence of rationals, and let $x$ be a real number. Show that if $a_{n} \leq x $ for all $n \geq 1 $, then $\lim_{n \rightarrow \infty} a_{n} \leq x $.

Answer 1

Since the concept of REAL LIMIT is not introduced until chapter 6, the (ε, N)-definition cannot be used to prove this. The hint is given to use contradiction and Prop 5.4.9, here's my proof using these two:

suppose $a=LIM_{n\to ∞}a_n>x$, there exists a rational $q$ that $a>q>x$ due to Prop 5.4.14. Since $\forall n≥1,a_n≤x<q$, construct a sequence $(q-a_n)$, it's trivial to prove the sequence is Cauchy and it's non-negative using Prop 5.4.9, that is, $$LIM_{n\to ∞}(q-a_n)=LIM_{n\to ∞}q-LIM_{n\to ∞}a_n=q-a≥0$$so $q≥a$, which contradicts $a>q$, deduced from the original hypothesis.

Answer 2

Since $(a_n)$ is Cauchy, it is converging to some $a\in\Bbb R$. Now, if $a>x$, then there exists $N$ such that $|a_N-a|<\frac{a-x}{2}$ and we get the contradiction: $$\frac{x-a}{2}<a_N-a \qquad \implies \qquad x <\frac{x+a}{2}<a_N$$