# prove: if a real number $a=\lim_{n\to \infty} a_n > c > 0$, then eventually $a_n > c$

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 $$\langle c_n \rangle$$.

• 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. Commented Jul 7, 2018 at 18:44
• 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. Commented Jul 7, 2018 at 18:47
• Oh! b.t.w;, that $c$ be rational or not is unimportant, I think. Commented Jul 7, 2018 at 18:49
• You're right; I simplified things a bit :) Commented Jul 7, 2018 at 18:54

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 $$-\dfrac{a-c}{2}, namely, $$a_n>a-\dfrac{a-c}{2}=\dfrac{a+c}{2}>c.$$ This is exactly what we want to prove.