0
$\begingroup$

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 :)

$\endgroup$
2
$\begingroup$

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 $a_n\in(-\infty,0).$

Added:

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Crosby Jul 4 '18 at 23:22
  • $\begingroup$ @Crosby If $a_j > \epsilon$ and $a_k<-\epsilon$ then $|a_j-a_k|>2\epsilon$. $\endgroup$ – David C. Ullrich Jul 4 '18 at 23:51
  • $\begingroup$ @Crosby: Ah! I see where (at least part of) your confusion lies. I will update my answer presently to address it. Hopefully, David's comment gets you the rest of the way, but I want to clarify anyway. $\endgroup$ – Cameron Buie Jul 4 '18 at 23:53
  • $\begingroup$ @CameronBuie (+1) Okay, thank you for the clarification! $\endgroup$ – Crosby Jul 5 '18 at 0:41
  • $\begingroup$ @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. $\endgroup$ – Crosby Jul 5 '18 at 13:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.