Proposition 5.4.9. Analysis I - Terence Tao.

Question

Proposition 5.4.9 (The non-negative reals are closed). Let $a_1, a_2, a_3, \ldots $ be a Cauchy sequence of non-negative rational numbers. Then $\text{LIM}_{n \to \infty}a_n$ is a non-negative real number.

Tao's Proof.

We argue by contradiction, and suppose that the real number $x:= \text{LIM}_{n \to \infty}a_n$ is a negative number. Then by definition of negative real number, we have $x:= \text{LIM}_{n \to \infty}b_n$ for some sequence $b_n$ which is negatively bounded away from zero, i.e, there is a negative rational $-c < 0$ such that $b_n \leq -c$ for all $n \geq 1$. On the other hand, we have $a_n \geq 0$ for all $n \geq 1$, by hypothesis. Thus the numbers $a_n$ and $b_n$ are never $c/2$-close, since $c/2 < c$. Thus the sequences $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ are not eventually $c/2$-close. Since $c/2 > 0$, this implies that $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ are not equivalent. But this contradicts the fact that both these sequences have $x$ as their formal limit.

My Question.

I was surprised to see Tao's proof was more complicated than mine. Where we differed was that I did not introduce a second sequence $(b_n)_{n=1}^{\infty}$. Why did he need to do this? Couldn't we just work with $(a_n)_{n=1}^{\infty}$? We would arrive at showing $a_n \leq -c < 0$ for all $n \geq 1$, which would contradict that every $a_n$ is non-negative.

Answer 1

In Chapter 5 Tao introduces the real numbers as formal expressions $\operatorname{LIM}_{n \to \infty} a_n$, where $(a_n)$ 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 $(a_n)$ and $(b_n)$ are equivalent Cauchy sequences. This means actually that $\operatorname{LIM}_{n \to \infty} a_n$ denotes the equivalence class of $(a_n)$ with respect to the equivalence relation "equivalent Cauchy sequences". See Confusion about Tao's construction of reals.

Thus $\operatorname{LIM}_{n \to \infty} a_n$ does not denote the "usual limit" of the sequence $(a_n)$. This concept is introduced only later in Chapter 6. The limit of a sequence $(x_n)$ of real numbers is written as $\lim_{n \to \infty} x_n$.

This explains why his proof uses the second sequence $(b_n)$. It is a different representative of the real number $x = \operatorname{LIM}_{n \to \infty} a_n$ and he shows that the assumption on $(b_n)$ leads to a contradiction.

Your proof comes too early for the status achieved in Chapter 5. Actually you work with $\lim_{n \to \infty} a_n$ and not with the formal expression $\operatorname{LIM}_{n \to \infty} a_n$.