0

What i'm proving is in the title. Essentially, I know that it holds for rational numbers. I want to prove it for all real numbers. The following is the definition for $a^x$ that I'll be using.

Let $\{r_n\}$ be any sequence of rationals that converge to $x$. Then, the exponential of the base $a$ is defined:

$$a^x = \lim_{n \to \infty} a^{r_n}$$

Proof Attempt:

Let $\{r_n\}$ be any sequence of rational numbers that converges to $x$.

This sequence is convergent, so it is bounded. By the Bolzano-Weierstrass Theorem, this has a convergent subsequence. We denote this subsequence by $\{s_n\}$.

This subsequence converges to $x$ as well. If it didn't, then there would exist an $\epsilon > 0$ such that for all $N>0$, $n>N$ and $|s_n-x| \geq \epsilon$. But since this subsequence is a part of the original sequence, which does converge to $x$, it follows that such an $\epsilon$ cannot exist.

If the subsequence is decreasing, then:

$$\forall n \in \mathbb{N}: s_n > 0$$

So, it follows that:

$$\forall n \in \mathbb{N}: a^{s_n} > 1$$

In the limit, it would follow that $a^x > 1$.

If the subsequence is increasing, then:

$$\exists N > 0: n > N \implies s_n > 0$$

$$\implies \exists N>0: n > N \implies a^{s_n} > 1$$

In the limit, it follows that $a^x > 1$. This proves the desired result.

Does the proof above work? If it doesn't, why? How can I fix it?

Mousedorff
  • 6,428
  • 2
    It doesn't work because $s_n=1/n\gt0$ but $a^{s_n}\to1$. – Peter Foreman Jun 10 '20 at 10:42
  • Thanks for the counterexample :D – Mousedorff Jun 10 '20 at 10:52
  • 1
    Wait hold on. That counterexample doesn't work, though? $s_n$ converges to 0 and it's a decreasing sequence. But $x > 0$ by hypothesis? Unless I'm completely misreading what you're saying haha – Mousedorff Jun 10 '20 at 10:55
  • Right @Peter but that's what I did? I did it in the last portion of the proof where I asserted the existence of $N$? – Mousedorff Jun 10 '20 at 10:57
  • I though about it again, and I am not sure anymore who is right. – Peter Jun 10 '20 at 11:03
  • Yea I'm not sure too. The counterexample given by Peter Foreman doesn't work as far as i know because the sequence he gave converges to 0, which is not the case here since $x > 0$. – Mousedorff Jun 10 '20 at 11:04
  • I think, the below answer is correct (and the counterexample as well). – Peter Jun 10 '20 at 11:12
  • Sure is that we cannot conclude that the limit is positive , if every entry from some point on is positive. But the below answer shows how to ensure that the limit is positive, and this is sufficient. – Peter Jun 10 '20 at 11:16
  • 1
    I mean, I'm pretty sure it is correct. I'm confident that the person who posted it wanted to post a correct proof. I'm just not sure if the counterexample provided works or not. That's what they're clarifiying in the comments haha. – Mousedorff Jun 10 '20 at 11:19
  • You don't need to invoke Bolzano-Weierstrass. You are starting with a convergent sequence, so any subsequence is automatically convergent to the same limit. What you do need is a theorem that allows you to pass to a monotonic subsequence (and you need to equip it with a rational lower bound, to avoid the kind of counterexample problem Peter Foreman mentioned). – Barry Cipra Jun 10 '20 at 11:22
  • But Peter Foreman's counterexample doesn't work as far as I can tell? Like I said, $\frac{1}{n} \to 0$ as $n \to \infty$ and so $x = 0$ in that case. But that goes against the hypothesis of the assertion. And well, the Monotone Subsequence Theorem shows that we can find a subsequence that's monotonic. That would guarantee that ${s_n}$, our chosen sequence, is monotonic and convergent. But yes, I should probably rewrite that first part. I don't really need Bolzano-Weierstrass, as you said. – Mousedorff Jun 10 '20 at 11:27
  • @AbhijeetVats, Peter's counterexample is directed at your logic when you say that if $s_n$ is a decreasing sequence of positive numbers, then from $a^{s_n}\gt1$ we can conclude that $\lim_{n\to\infty}a^{s_n}\gt1$. As I mentioned parenthetically, you need a rational lower bound on the sequence in order to guarantee strict inequality in the limit. But that's not hard to do: Just pick a rational number $r$ between $0$ and $x$, so if $s_n$ decreases to $x$, then $s_n\gt r$ for all $n$, so $a^{s_n}\gt a^r$ and thus $\lim_{n\to\infty}a^{s_n}\ge a^r\gt1$. – Barry Cipra Jun 10 '20 at 11:37
  • Ohhhh okay that works, then. I didn't really understand what he was directing it towards, which is why I believed it to not work. – Mousedorff Jun 10 '20 at 11:44

1 Answers1

2

It doesn't work, because you merely proved that $a^{s_n}\ge1$, if we assume the map is continuous.

To fix it, we'll assume the following exponentiation fact:

  • $a^q$ is increasing for rational $q$, $a>1$
  • $a^q$ is greater than $1$ for for positive rational $q$, $a>1$

Consider any sequence $r_n$ that converges to $x$. Then we know that for $\varepsilon=\frac x2$, eventually every term in the sequence $a^{r_n}$ is greater than $1+\kappa$ for $\kappa=a^{\frac x2}-1>0$.

Thus, since eventually every term is greater than $1+\kappa$, the limit is at least $1+\kappa$ (note that it could be equal to $1+\kappa$, but this is still $>1$).

Of course, if you haven't already done so, you also need to show that this limit exists and is well-defined (independent of which sequence you choose).

hdighfan
  • 4,205
  • Oh nice. I've already proved those facts that you used so I'll just use them to fix the argument. Just to be sure, you're saying that the mistake in my reasoning occurs in the last few lines of the argument, right? Also, yes, I've already shown that the limit exists and is well-defined independent of the chosen sequence. – Mousedorff Jun 10 '20 at 10:52
  • 1
    The mistake comes when you say that $a^{s_n}>1$ for all $n$ implies that in the limit, $a^x>1$ is true as well -- what happens if $a^{s_n}=1+\frac1n$ or some other dumb sequence like that? Then, sure, each term is greater than $1$, but the limit isn't. However, if you set up a buffer and prove $a^{s_n}>1+\kappa$ for some fixed $\kappa>0$, then you're sweet (if you look at my definition of $\kappa$, it's clearly independent of $n$, which is the important part). – hdighfan Jun 10 '20 at 11:05
  • I converted my downvote to an upvote, since I missed this part. It is exactly the way to solve the problem. – Peter Jun 10 '20 at 11:11
  • I understand the argument in the later parts of your comment. That makes sense. But, well, if you assert that $a^{s_n} = 1+\frac{1}{n}$, then we can say that $s_n = \log_a(1+\frac{1}{n})$ and this goes to 0 when $n \to \infty$. Of course, we're speaking very informally here because the logarithm hasn't been defined but if it was, then that would be the result, no? So, it wouldn't fulfill the correct conditions to be the required counterexample? Unless the issue with what I'm saying is that $s_n$ isn't necessarily the logarithm of $1+\frac{1}{n}$? – Mousedorff Jun 10 '20 at 11:13
  • Also, sorry if it seems like I'm dissing what you're saying haha. I'm just making sure I'm getting every part of this absolutely correct. – Mousedorff Jun 10 '20 at 11:14
  • 2
    Nah, it's fine. That's correct too -- if the sequence really were $1+\frac1n$, then $s_n\to0$ is what would happen. In other words, if $a^{s_n}\to1$ then $s_n\to0$ is intuitively true. (and, indeed, if it weren't true, then the question would be false)

    The problem is that you haven't actually proved that, and proving that that's true is essentially doing logarithms (which I assumed would be covered after arbitrary real powers :)). So yeah, it's not actually a counterexample, but the main issue is your proof doesn't explicitly rule out that case.

     – hdighfan Jun 10 '20 at 11:15
  • Okay yes, that makes a lot of sense. Thanks so much for the assistance :D – Mousedorff Jun 10 '20 at 11:24