Prove that for all $x\in\Bbb R$, precisely one of the statements $x>0$, $x=0$, or $x<0$ holds

I came across the following exercise while studying Terrence Tao's book Analysis I:

Exercise 5.4.1 Let $x\in\Bbb R$. Show that precisely one of the statements $x>0$, $x=0$, or $x<0$ holds

My Attempt: First let us prove that at least one of these statements must hold. Suppose $x=\operatorname{LIM}_{n\to\infty}a_n$ is a nonzero real number (where $\operatorname{LIM}_{n\to\infty}$ denotes the formal limit of the sequence that follows). Then the sequence $(a_n)_{n\in\Bbb N}$ cannot be equivalent to $(0)_{n\in\Bbb N}$, i.e. $$\neg(\forall\varepsilon >0 \ \exists N\in\Bbb N \ \forall n\ge N:|a_n|\le\varepsilon)\equiv \exists\varepsilon >0 \ \forall N\in\Bbb N \ \exists n\ge N:|a_n|>\varepsilon.$$ The author hints to conclude that $(a_n)_{n\in\Bbb N}$ is either eventually positively or negatively bounded away from zero; namely $$\exists c>0 \ \exists N\in\Bbb N \ \forall n\ge N: |a_n|\ge c \quad \lor \quad \exists c<0 \ \exists N\in\Bbb N \ \forall n\ge N: |a_n|\le c.$$ Clearly, there are infinitely many $n\in\Bbb N$ such that $|a_n|>\varepsilon$, but there is no certainty that this holds for all $a_n$. Nor is there any certainty that the claim holds for all $n\ge N$ for some $N$. So how can I conclude? I have no problems showing that only one of these can hold at once.

Thank you for your time, and happy 4th for those who celebrate :)

Well, remember that the sequences in question are Cauchy, so since there are infinitely-many $a_n\in(-\infty,-\varepsilon)\cup(\varepsilon,\infty),$ we can actually show that either (i) there are all but finitely-many $a_n\in(0,\infty)$ or (ii) there are all but finitely-many $a_n\in(-\infty,0).$

One difference in approach that I would take is to say that $$\neg\left(\forall\varepsilon>0,\exists N\in\Bbb N:\forall n\ge N,\left|a_n\right|\le\varepsilon\right)\equiv\exists\varepsilon_0>0:\forall N\in\Bbb N,\exists n\ge N:|a_n|>\varepsilon_0.$$
Aside from some fiddly notational stuff (like comma usage and colon placement), the big change is using $\varepsilon_0$ to denote a particular value of $\varepsilon.$ Since we're still hoping to prove something that holds for all $\varepsilon,$ it behooves us to make clear which one that is.
Now, use the definition of Cauchy sequence to show that there exists some $N_0\in\Bbb N$ such that for all $m,n\ge N_0,$ we have $|a_m-a_n|<\varepsilon_0.$ We already know that there is some $n\ge N)$ such that $\left|a_n\right|>\varepsilon_0.$ From this, we can show that for all $m\ge N_0,$ we must have $a_m\in\left(a_n-\varepsilon_0,a_n+\epsilon_0\right),$ and since $|a_n|>\epsilon_0,$ then we're done.
• I'm not sure I quite understand. How does $(a_n)_{n\in\Bbb N}$ being a Cauchy sequence help? I suppose that you could say that there are all but finitely many $a_n\in (\varepsilon, \infty)$ or all but finitely many $a_n\in (-\infty, \varepsilon)$ for some $\varepsilon$. If it held for all epsilon, then I would understand. Jul 4, 2018 at 23:22
• @Crosby If $a_j > \epsilon$ and $a_k<-\epsilon$ then $|a_j-a_k|>2\epsilon$. Jul 4, 2018 at 23:51
• @CameronBuie I probably should’ve asked this before I accepted your answer, and I’m sorry to bother you again, but if $(a_n)_{n\in\Bbb N}$ is eventually positively bounded away from zero or eventually negatively bounded away from zero, then why should $x$ be positive or negative? The word eventually is what confuses me here. Does it suffice to prove that the sequences $(a_n)_{n\in\Bbb N}$ and $(a_n)_{n=N}^\infty$ are equivalent? Everything else is crystal clear, though. Jul 5, 2018 at 13:47