# Exercise 6.7.1 of Tao’s Analysis I

In Exercise 6.7.1 of Analysis I, Tao asks (among several things) to prove that for a real number $$x>0$$ and real numbers $$\alpha$$ and $$\beta$$, $$(x^\alpha)^\beta=x^{\alpha\beta}$$.

I’m stuck because following his Definition 6.7.2, I can write (I’ve proved this) $$(x^\alpha)^\beta=\lim\limits_{n’\to\infty}\lim\limits_{n\to\infty} x^{q_n r_{n’}}$$ where $$(q_n)_{n=1}^\infty$$ and $$(r_n)_{n=1}^\infty$$ are rational sequences converging to $$\alpha$$ and $$\beta$$ respectively.

But Tao doesn’t mention anywhere before this section how to handle double limits, and I have no clue how to prove that this is equal to $$\lim\limits_{n\to\infty} x^{q_n r_n}$$ which is equal to $$x^{\alpha\beta}$$.

Maybe I’m on the wrong track. Any help would be great.

You need Lemma 6.7.1 (p152) "Continuity of exponentiation" when he proves that the result of real exponentiation is independent of the sequence (of rationals) converging to $$a$$ (only the limit counts).\ First, remark that, if $$\alpha$$ or $$\beta$$ is zero, then $$(x^\alpha)^\beta=x^{\alpha\beta}$$ holds.

Then, we first suppose $$\alpha>0$$ (resp. $$\beta>0$$) and replace $$q_n$$ (resp. $$r_m$$) by any increasing sequence of strictly positive rationals $$\hat{q}_n$$ (resp. $$\hat{r}_m$$) converging to $$\alpha$$ (resp. $$\beta$$).

Due to the fact that the sequences are increasing, we have, for all $$n,m$$ $$x^{\hat{q}_{n}\cdot \hat{r}_{m}}=(x^{\hat{q}_{n}})^{\hat{r}_{m}}\leq (x^{\alpha})^{\beta}\ ;\ (x^{\hat{q}_{n}})^{\hat{r}_{m}}=x^{\hat{q}_{n}\cdot \hat{r}_{m}}\leq x^{\alpha\cdot \beta}$$ in the first one you take the limit $$n\to\infty$$ and get $$x^{\alpha\cdot \hat{r}_{m}}\leq (x^{\alpha})^{\beta}$$ then, with $$m\to\infty$$, you get $$x^{\alpha\cdot \beta}\leq (x^{\alpha})^{\beta}$$. Using the second one, taking the limits in the same order ($$n$$ then $$m$$) you get successively $$(x^{\alpha})^{\hat{r}_{m}}\leq x^{\alpha\cdot \beta}$$ and then $$(x^{\alpha})^{\beta}\leq x^{\alpha\cdot \beta}$$.

You finish the cases, using $$x^{-a}=\dfrac{1}{x^{a}}$$.

Hope it helps !

• I'm not getting what you say in the second point. The LHS doesn't look correct.
– Atom
Apr 10, 2020 at 16:22
• Further, I don't think any notion of $\text{sup}_{n,n'}$ is defined for two variables -- $n$ and $n'$.
– Atom
Apr 10, 2020 at 16:32
• Sorry, but I don’t understand...
– Atom
Apr 10, 2020 at 17:09
• OK, I'll find another way, let me some time (I am partly busy :) Apr 10, 2020 at 17:26
• Many thanks to you!
– Atom
Apr 10, 2020 at 17:32

For anyone reading this, I have managed to get a more satisfying proof (imo). It is very long though!

1. First show that for any $$x>0$$ and $$\alpha > \beta$$, we have $$x^\alpha \ge x^\beta$$ if $$x\ge 1$$ and $$x^\alpha \le x^\beta$$ if $$x < 1$$. This is the easy part!
2. Next, show that if $$(a_n)_{n=1}^\infty$$ is a sequence of real numbers converging to $$a$$ then $$(x^{a_n})_{n=1}^\infty$$ converges to $$x^a$$ (Hint: consider approximating $$(a_n)_{n=1}^\infty$$ by a sequence of rational numbers and using 1).

Let $$q_n := \frac{\lfloor 2^na_n \rfloor}{2^n}$$ for each $$n\ge 1$$ (you might want to review Exercise 5.4.3), then $$(q_n)_{n=1}^\infty$$ is a sequence of rational numbers satisfying the inequality $$0 \le a_n - q_n < 1/2^n$$. Thus by the squeeze test, $$\lim_{n\to \infty} q_n = a$$ so by definition, $$x^a = \lim_{n\to \infty} x^{q_n}$$. Rearranging the previous inequality we get, $$q_n - 1/2^n < a_n < q_n + 1/2^n$$, so using 1 we get $$x^{q_n}x^{-1/2^n} \le x^{a_n} \le x^{q_n}x^{1/2^n}$$ (or with the inequalities flipped, but that doesn't matter). Using the squeeze test once more, we get the desired result.

1. Now prove that if $$(a_n)_{n=1}^\infty$$ is sequence of positive real numbers which converges to some $$L > 0$$, then $$(a_n^R)_{n=1}^\infty$$ converges to $$L^R$$ for any rational number $$R$$, i.e $$\lim_{n\to \infty}a_n^R = (\lim_{n\to \infty}a_n)^R$$ (Hint: prove this for the case $$R=a$$ and $$R=1/b$$ where $$a,b \in \mathbb{Z}$$ and $$b\ne 0$$. Then combine these together. For the second case, you might want to "normalize" $$a_n$$ to make it converge to $$1$$).

It's easy to show that $$\lim_{n\to \infty}a_n^a = L^a$$ for any integer $$a$$ using the limit laws and induction. Now let $$b > 0$$ be an integer, we want to show that $$(a_n^{1/b})_{n=1}^\infty$$ converges to $$L^{1/b}$$. Using the limit laws, this is equivalent to showing that $$(y_n^{1/b})_{n=1}^\infty$$, where $$y_n := \frac{a_n}{L}$$, converges to $$1$$. To this end, we need to show that for every $$\varepsilon > 0$$, $$(y_n^{1/b})_{n=1}^\infty$$ is eventually $$\varepsilon$$-close to $$1$$; also we may assume that $$\varepsilon < 1$$ (Why?). Thus we need some $$N \ge 1$$ such that $$1-\varepsilon \le y_n^{1/b} \le 1 + \varepsilon$$ for all $$n \ge N$$, that is by Lemma 5.6.9, $$(1-\varepsilon)^b \le y_n \le (1 + \varepsilon)^b$$ for all $$n \ge N$$. This looks like progress! Since $$\lim_{n\to \infty}y_n = 1$$ we may pick any $$\varepsilon' > 0$$ so that $$1-\varepsilon' \le y_n \le 1 + \varepsilon'$$, for sufficiently large $$n$$. Thus, it is sufficient to require that eventually, $$(1-\varepsilon)^b \le 1-\varepsilon' \le y_n \le 1+\varepsilon' \le (1+\varepsilon)^b$$. In other words, we need both of $$\varepsilon' \le (1+\varepsilon)^b-1$$ and $$\varepsilon' \le 1 - (1-\varepsilon)^b$$ to hold. This is easily achieved by picking $$\varepsilon' := \min((1+\varepsilon)^b-1, 1 - (1-\varepsilon)^b)$$ which is positive (Why?). To finish this part off, let $$R := a/b$$ where $$a$$ and $$b$$ are integers and $$b>0$$ then we know that $$(a_n^{1/b})_{n=1}^\infty$$ converges to $$L^{1/b}$$ and of coures, $$a_n^{1/b}, L^{1/b} > 0$$ for all $$n\ge 1$$, so $$((a_n^{1/b})^a)_{n=1}^\infty$$ converges to $$(L^{1/b})^a = L^R$$.

4.Finally, using 2 and 3 show that for every $$x>0$$ and all real numbers $$q$$ and $$r$$, $$(x^q)^r = x^{qr}$$ (Hint: first prove this for rational $$r$$).

Let $$(q_n)_{n=1}^\infty$$ and $$(r_n)_{n=1}^\infty$$ be two sequences of rational numbers converging to $$q$$ and $$r$$ respectively. First assume that r is rational, then by 3 we have $$(x^q)^r = (\lim_{n\to \infty}x^{q_n})^r = \lim_{n\to \infty}(x^{q_n})^r = \lim_{n\to \infty}x^{q_nr}$$. Since $$q_nr \to qr$$ as $$n \to \infty$$, 2 guarantees that this last limit is $$x^{qr}$$, although we don't actually need 2 here (Why?). Now we prove the general case: since each $$r_n$$ is rational we have, $$(x^q)^r = \lim_{n\to \infty}(x^q)^{r_n} = \lim_{n\to \infty}x^{qr_n}$$. This time we need 2 to conclude that this limit is $$x^{qr}$$ as it should be.

Notes: I am just a self-learner so there might be some mistakes or subtleties that I didn't notice. Also, some of my proofs might feel out of the blue, here is some of my thought process: When it comes to unsolvable-looking problems like this one, I try to prove as many simple statements which could be related to it. While tinkering around with it, I noticed that I needed part 2 and 3 above. The way I came up with 2 is as follows: I want a sequence of rational numbers and each term must satisfy two things, first it needs to be fully determined by the $$a_n$$'s and second, it needs to get closer and closer to the $$a_n$$'s. I know one function which can give a unique rational number for each real number, the floor function $$\lfloor\cdot\rfloor$$. To get it closer I scaled $$x$$ up by $$2^n$$ and then scaled back down, if this makes any sense. Part 3 was the most difficult for me, I had to play around with it for a while before I got the idea of simplifying the problem, by getting rid of $$L^{1/b}$$. All in all, it took me an embarrassingly long time to finish this proof. It was kind of frustrating, since Terence Tao said "the other parts are similar", what? I'm I missing something because just like you, I figured it is futile to work with that double limit using my limited knowledge (see what I did here).