As a clarification, I was working through the construction of real numbers in the Analysis I textbook by Terrence Tao. At this point, real numbers have just been defined as the ``formal'' limits (i.e., equivalence classes) of Cauchy sequences of rational numbers; so the $\lim_{n\to \infty} a_n $ in question (and the rest of this post) is only a notation for the equivalence class of a Cauchy sequence of rantionals $\langle a_n \rangle$, and doesn't have the usual $\epsilon-\delta$ definition (which is not yet introduced).

Here's my attempt at proving the statement in question. I begin with a lemma:

Lemma: If a real number $x = \lim_{n\to \infty} x_n > 0$, then the terms of the Cauchy sequence $\langle x_n \rangle$ must be eventually positive, i.e., there exists $N \in \mathbb{N}$ s.t., $\forall n\geq N$, $x_n >0$.

Proof: By definition of $x>0$, $x = \lim_{n\to \infty} y_n$ for some $\langle y_n \rangle $ that is positively bounded away from zero, so that $\forall n\geq N, y_n \geq w$, where $w>0 \in \mathbb{Q}$. Since $x =\lim_{n\to \infty} x_n = \lim_{n\to \infty} y_n$, the Cauchy sequences $\langle x_n \rangle$ and $\langle y_n \rangle$ are equivalent, then if we pick $\epsilon = \frac{w}{2}>0$, there exists $N$ s.t. $\forall n \geq N, |x_n - y_n| \leq \frac{w}{2}$, which implies $x_n - y_n \geq -\frac{w}{2}$; adding $y_n \geq w$ to both sides gives $x_n \geq \frac{w}{2} > 0$.

Now for the main proof:

Since $a > c$, the formal limit of the Cauchy sequence of rationals $\langle a_n - c\rangle$ must be positive. By definition, this means $\lim_{n\to \infty} a_n -c = \lim_{n\to \infty} b_n$ for some Cauchy sequence $\langle b_n \rangle$ that is positively bounded away from zero, i.e., $\exists d >0, \forall n, b_n \geq d$. By definition of equality of real numbers, the sequences $\langle a_n -c \rangle$ and $\langle b_n \rangle$ are equivalent, so for every $\epsilon >0$, there exists $N_1 \in \mathbb{N}$ s.t. $\forall n \geq N_1$, $|a_n - c - b_n| \leq \epsilon$.

Since $a>0$, by our lemma there exists $N_2$ such that $\forall n\geq N_2, a_n > 0$. Let $N = \max \{N_1, N_2\}$.

Now assume for contradiction that it's not the case that eventually $a_n > c$; then there must exist $n_0\geq N$ with $a_{n_0} \leq c$, for otherwise eventually $\forall n\geq N, a_n > c$, a contradiction.

Then we have

$$c + d - a_{n_0} \leq c + b_n- a_{n_0} = |c + b_n| - |a_{n_0}| \leq |a_{n_0} - c - b_n| \leq \epsilon $$ so $a_{n_0} \geq c + d - \epsilon$.

If we choose $\epsilon=d/2$, then $a_{n_0} \geq c + d/2$, contradicting our assumption that $a_{n_0} \leq c$.

Is this correct? Any simpler or alternative proof is also appreciated.

[Update]: Looking at this again, it seems a direct proof would have been simpler: Let $\langle a_n \rangle$, $\langle c_n \rangle$ be any two Cauchy sequences of rationals such that $a = \text{LIM}_{n\to\infty}a_n$ and $c = \text{LIM}_{n\to\infty}c_n$. Since $a-c>0$ is equivalent to some Cauchy sequence $\langle b_n \rangle$ bounded away from 0, there exists some rational $q >0$ s.t. eventually $ a_n - c_n \geq q > 0$ (this actually follows from my lemma, with $q=\frac{w}{2}$). Then it shouldn't be hard to show that eventually $a_n > c$, using the fact that $c$ is the equivalent class of a Cauchy sequence $\langle c_n \rangle$.

  • $\begingroup$ In the original post I intentionally used $lim$ instead of $\lim$, to emphasize that the real numbers are taken to be the formal (instead of "genuine") limits of rational numbers, i.e. a real $x$ is an object of the form $lim_{n\to\infty} x_n$ for some Cauchy sequence of rationals. But I guess the difference isn't critical. $\endgroup$
    – Yibo Yang
    Jul 7, 2018 at 18:44
  • $\begingroup$ I edited your for this problem. Very often, it isn't intentional, so when I happen to be looking at a post with this problem, iItake the liberty of editing it. Anyway, I don't think the difference is important here. $\endgroup$
    – Bernard
    Jul 7, 2018 at 18:47
  • $\begingroup$ Oh! b.t.w;, that $c$ be rational or not is unimportant, I think. $\endgroup$
    – Bernard
    Jul 7, 2018 at 18:49
  • $\begingroup$ You're right; I simplified things a bit :) $\endgroup$
    – Yibo Yang
    Jul 7, 2018 at 18:54

1 Answer 1



Let $\varepsilon=\dfrac{a-c}{2}>0.$ Then, accoring to the $\varepsilon-N$ definition of the sequence limit, we have that,for the $\varepsilon=\dfrac{a-c}{2}>0,$ there exists a $N \in \mathbb{N_+}$ such that $|a_n-a|<\varepsilon=\dfrac{a-c}{2}$when $n>N$. Thus, when $n>N$, we may have $-\dfrac{a-c}{2}<a_n-a$, namely, $a_n>a-\dfrac{a-c}{2}=\dfrac{a+c}{2}>c.$ This is exactly what we want to prove.

  • 2
    $\begingroup$ That's for ordinary convergence, isn't it? As I understood the question it's a Cauchy sequence, not necessarily converging in the ordinary sense. $\endgroup$
    – md2perpe
    Jul 7, 2018 at 19:05
  • $\begingroup$ The Real Numbers with the Euclidan Normal is a Complete Metric Space. This means a sequence is Cauchy for the reals Number with the euclidan normal IF and ONLY IF, it is convergent. $\endgroup$
    – user518441
    Jul 7, 2018 at 19:46
  • $\begingroup$ I know, but since Yibo Young started to talk about Cauchy sequences it seemed like we should work with such. $\endgroup$
    – md2perpe
    Jul 7, 2018 at 20:21

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