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.


2 Answers 2


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 !

  • $\begingroup$ I'm not getting what you say in the second point. The LHS doesn't look correct. $\endgroup$
    – Atom
    Apr 10, 2020 at 16:22
  • $\begingroup$ Further, I don't think any notion of $\text{sup}_{n,n'}$ is defined for two variables -- $n$ and $n'$. $\endgroup$
    – Atom
    Apr 10, 2020 at 16:32
  • $\begingroup$ Sorry, but I don’t understand... $\endgroup$
    – Atom
    Apr 10, 2020 at 17:09
  • 1
    $\begingroup$ OK, I'll find another way, let me some time (I am partly busy :) $\endgroup$ Apr 10, 2020 at 17:26
  • $\begingroup$ Many thanks to you! $\endgroup$
    – 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).


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