Terence Tao Exercise 5.4.8: Boundedness of Limit. 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 $. 
 A: 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.
A: 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:
$$\frac{x-a}{2}<a_N-a \qquad \implies \qquad x <\frac{x+a}{2}<a_N$$
A: I like @bigtit answer. However, the question says the solution could be arrived at using Corollary 5.4.10, and this is my attempt using this corollary.
Let's proceed by contradiction and prove the statement: 
$$ \forall n \ge1, a_n \le x \Rightarrow LIM_{n\to \infty} a_n > x  $$
Since $x\in \Bbb R$ and $a = LIM_{n\to \infty} a_n  $ we know there exists $ q\in \Bbb Q  $ such that  $x < q < a$. By hypothesis $$ \forall n \ge1, q > x \Rightarrow q>a_n $$ Since $q \in \Bbb Q$ it can also be expressed as the Cauchy sequence $ LIM_{n\to \infty} q_n $. Applying Corollary 5.4.10: 
$$ \forall n \ge1,  q_n> a_n \Rightarrow LIM_{n\to \infty} q_n >  LIM_{n\to \infty} a_n$$ 
Thus $q = q_n > LIM_{n\to \infty} a_n $, a contradiction. Thus our initial statement is false.
