Am asking in the hopes that someone has read Tao's book and can explain his proof outline; this question comes from Tao's Analysis 1. The proposition being proved is:
Proposition 6.1.15 (Formal limits are genuine limits). Suppose that $(a_n)_{n=1}^\infty$ is a Cauchy sequence of rational numbers. Then $(a_n)_{n=1}^\infty$ converges to $\operatorname{LIM}_{n\to\infty}a_n$, i.e. $$\operatorname{LIM}_{n\to\infty}a_n=\lim_{n\to\infty}a_n.$$
The outline he provides is as follows, with the bolded part being the part I do not understand:
Let $(a_n)_{n=m}^\infty$ be a Cauchy sequence of rationals, and write $L:=\operatorname{LIM}_{n\to\infty}a_n$. We have to show that $(a_n)_{n=m}^\infty$ converges to $L$. Let $\varepsilon>0$. Assume for the sake of contradiction that sequence $a_n$ is not eventually $\varepsilon$-close to $L$. Use this, and the fact that $(a_n)_{n=m}^\infty$ is Cauchy, to show that there is an $N\geq m$ such that either $a_n>L+\frac{\varepsilon}{2}$ for all $n\geq N$, or $a_n>L-\frac{\varepsilon}{2}$ for all $n\geq N$. Then use Exercise 5.4.8.
My issue with his proof outline is that I do not see where contradiction comes into play. We are essentially showing that there exists some $N$ such that for all $n\geq N$, we have $|a_n-L|<\frac\varepsilon2$, in other words, that $(a_n)$ converges to $L$. This can be shown from the fact that $(a_n)$ is Cauchy, as all Cauchy sequences converge. I would greatly appreciate if someone could explain how Tao's proof is supposed to work, thank you.
Potentially relevant definitions:
Definition 5.3.1 (Real numbers). A real number is defined to be an object of the form $\operatorname{LIM}_{n\to\infty}a_n$, where $(a_n)_{n=1}^\infty$ is a Cauchy sequence of rational numbers. Two real numbers $\operatorname{LIM}_{n\to\infty}a_n$ and $\operatorname{LIM}_{n\to\infty}b_n$ are said to be equal iff $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are equivalent Cauchy sequences.
Definition 6.1.5 (Convergence of sequences). Let $\varepsilon>0$ be a real number, and let $L$ be a real number. A sequence $(a_n)_{n=N}^\infty$ of real numbers is said to be $\varepsilon$-close to $L$ iff $a_n$ is $\varepsilon$-close to $L$ for every $n\geq N$, i.e., we have $|a_n-L|\leq\varepsilon$ for every $n\geq N$. We say that a sequence $(a_n)_{n=m}^\infty$ is eventually $\varepsilon$-close to $L$ iff there exists an $N\geq m$ such that $(a_n)_{n=N}^\infty$ is $\varepsilon$-close to $L$. We say that a sequence $(a_n)_{n=m}^\infty$ converges to $L$ iff it is eventually $\varepsilon$-close to $L$ for every real $\varepsilon>0$.
Definition 6.1.8 (Limits of sequences). If a sequence $(a_n)_{n=m}^\infty$ converges to some real number $L$, we say that $(a_n)_{n=m}^\infty$ is convergent and that its limit is $L$; we write $$L=\lim_{n\to\infty}a_n$$ to denote this fact.
Best Answer
The consequence of the assumption in the bold-faced text is the following:
So now there are two possible cases arising from the assumption:
Case $a_n>L+\frac{\varepsilon}{2}$ for all $n\geq N$. Then you can use Exercise 5.4.8 (possibly after some renaming/renumbering of the relevant sequence so it matches the conditions of the exercise) to show that $$ \operatorname{LIM}_{n\to\infty}a_n \geq L+\frac{\varepsilon}{2}, $$ contradicting the premise that $L:=\operatorname{LIM}_{n\to\infty}a_n.$
Case $a_n>L-\frac{\varepsilon}{2}.$ Then you can use Exercise 5.4.8 to show that $$ \operatorname{LIM}_{n\to\infty}a_n \leq L-\frac{\varepsilon}{2}, $$ again contradicting the premise that $L:=\operatorname{LIM}_{n\to\infty}a_n.$
Since you get a contradiction either way, the assumption leads to a contradiction.
Perhaps you can make a direct proof. But this one does indeed work by contradiction.