Irrational Exponents - Lebl's Real Analysis 1.2.17

Question

I am working on a portion of part (d) from problem 1.2.17 in Lebl's Introduction to Real Analysis (available here).

Problem Statement:

For an irrational $z \in \mathbb{R} \setminus \mathbb{Q}$ and $x > 1$, define:

$ x^z := \sup\{x^r : r \leq z, r \in \mathbb{Q}\}. $

I want to prove the forward implication $x < y \implies x^z < y^z$ for $x > 1$ and $y > 1$.

For brevity, let's denote the set $\{x^r : r \leq z, r \in \mathbb{Q}\}$ as $\{x^r\}$.

Background:

It has been established in a previous exercise that this statement holds true for positive rational exponents.

Proof Sketch:

1. Observation: Note that $x^z \notin \{x^r\}$ since for any $x^r \in \{x^r\}$, there exists some $x^s > x^r$ with $r < s < z$ due to the density of rationals in the reals.

2. Comparison: For every $a \in \{x^r\}$, there exists $b \in \{y^r\}$ such that $a < b < \sup\{y^r\} = y^z$. Thus, $y^z$ is an upper bound for $\{x^r\}$.

   Therefore, we conclude that:

   $\sup\{x^r\} \leq \sup\{y^r\}$,

   or $x^z \leq y^z$.

Challenge:

I am encountering difficulty in demonstrating that $x^z \neq y^z$. While it is intuitive, and given the absence of logarithms in the text up to this point, it is assumed that a proof using properties of real numbers and suprema is sufficient.

Goal:

It would suffice to show that there exists some $y^r$ that is not in $\{x^r\}$, or equivalently, find $\epsilon > 0$ such that:

$ y^r \geq x^z + \epsilon $

for some $0 < r < z$.

Any insights or guidance on constructing this part of the proof would be greatly appreciated. Or maybe I am just overthinking this entirely.

Thank you!

Answer 1

Here are some relevant exercises from Lebl's text I used as the basis of the proof.

Exercise 1.1.9

Let $ S $ be an ordered set and $ A $ is a nonempty subset such that $\sup A$ exists. Suppose there is a $ B \subset A $ such that whenever $ x \in A $ there is a $ y \in B $ such that $ x \leq y $. Show that $\sup B$ exists and $\sup B = \sup A$.

Proposition 1.2.6

Let $ A \subset \mathbb{R} $ be nonempty.

(i) If $ x \in \mathbb{R} $ and $ A $ is bounded above, then $\sup(x + A) = x + \sup A$.

Lebl defines $x + A$ := $\{x + y \in \mathbb{R}: y \in A \subset \mathbb{R}\} $.

Proof

Again, for brevity, denote the set $\{x^r : r \leq z, r \in \mathbb{Q}\}$ as $\{x^r\}$

To prove that $x < y \implies x^z < y^z $ for $1 < x, y$ (real numbers) and irrational $z > 0$, consider the following lemmas:

Lemma 1

If $0 < r < s$ for rationals $r$ and $s$, and $1 < x, y$ (real numbers), then $y^s - x^s > y^r - x^r$.

Proof: Since $s = r + t$ for some rational $t > 0$,

$$ \begin{align*} y^s - x^s - (y^r - x^r) &= y^{r+t} - x^{r+t} - y^r + x^r \\ &= y^r y^t - x^r x^t - y^r + x^r \\ &= y^r(y^t - 1) - x^r(x^t - 1) \\ &> y^r(y^t - 1) - x^r(y^t - 1) \\ &= (y^t - 1)(y^r - x^r) \\ &> 1 \cdot (y^r - x^r) \quad \text{since } y^t > 1 \\ &= y^r - x^r > 0. \end{align*} $$

Thus, $y^s - x^s > y^r - x^r$.

Lemma 2

For rationals $r < s < z$ (with $z$ irrational), $\sup \{x^r\} = \sup \{x^s\}$ for $1 < x$.

Proof: By Exercise 1.1.9, if $\{x^s\}$ is a proper subset of $\{x^r\}$, and for every element of $\{x^r\}$ there is an element of $\{x^s\}$ that is larger, $\sup \{x^s\} = \sup \{x^r\}$. Hence, both suprema equal $x^z$.

Lemma 3

Using Proposition 1.2.6 (i): For real $x$ and set $A$ bounded above, $\sup(x + A) = x + \sup A$.

Main Proof

From Lemma 1, for $0 < r < s < z$ (where $r$ and $s$ are rationals), let $\epsilon = y^r - x^r$. Thus, $y^s > x^s + \epsilon$.

From Lemma 2, $\sup\{x^s\} = \sup\{x^r\} = x^z$. Similarly, $\sup\{y^s\} = \sup\{y^r\} = y^z$.

Since each element of $\{x^s\} + \epsilon$ is less than the corresponding element in $\{y^s\}$, $\sup\{y^s\}$ is an upper bound for $\{x^s\} + \epsilon$.

Thus, $$ \sup(\{x^s\} + \epsilon) \leq \sup\{y^s\} = y^z. $$

By Lemma 3, $$ \sup(\{x^s\} + \epsilon) = \sup\{x^s\} + \epsilon = \sup\{x^r\} + \epsilon = x^z + \epsilon. $$

Therefore, $x^z < x^z + \epsilon \leq y^z$, as desired. $\blacksquare$