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The outline of the exercise is that: Fix $b > 1, y > 0$ and show that there is a unique real $x$ such that $b^x = y$. (For further specification see e.g. this post) Part f.) requires to show that $b^x = y$ where $x = \sup A, A = \{w \in \mathbb{R}\mid b^w < y\}$. It is very well possible that I am too pedantic, but I am struggling to prove the part since I can't even show that $A$ is necessarily 1.) non-empty 2.) bounded above.

The three cases to consider are that i.) $b < y$, ii.) $b = y$, iii.) $b > y$. In the second case the claim is straightforward to prove. But how can you show in the i.) and iii.) that $A$ is bounded above and non-empty, respectively? Specifically I am stuck at the thought that since we are proving the existence of the logarithm, and at this point in the book the author hasn't covered sequences, series and limits, you can't really make an argument that you can multiply/divide $b$ enough times by itself to get a value greater/less than $y$.

Epsilon Away
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    Prove (using the exercise preceding this one in the book which covers the basic results about $b^x$) the fact that if $b > 1$, then $b^w$ is an increasing function of $w$. The positive integers powers $b^n$ are unbounded above if $b > 1$ so they must eventually exceed $y$. Can you conclude from that that the set $A$ which you describe is bounded above? – shoteyes Jun 06 '21 at 08:17
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    When I say eventually, I want to clarify that this does not directly refer to a limit of any kind. Since part (a) of the exercise tells us that $b^n \geq n (b - 1) + 1$ when $b > 1$ and $n$ is a positive integer, the fact that the set ${b^n: n \in \mathbb{N}}$ does not have an upper bound follows from the inequality and the Archimedean property. – shoteyes Jun 06 '21 at 08:28
  • @shoteyes I manage to prove the case when $b < y$ by using the Archimedean property. Do you have tips on how to show the case when $y < b$? My initial thought was to apply Achimedean property again so that $yk > b, k \in \mathbb{N}$, and then use the part b.), so that $yk > b \geq k(b^{1/n} - 1) \Longleftrightarrow y > b^{1/n} - 1$. But this construction only shows that $b^{1/n} - 1$ is less than $y$ for some $k$, while I would need that $y > b^{1/k}$. – Epsilon Away Jun 06 '21 at 10:58

2 Answers2

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I’m expanding upon my comments for this answer.

I will use the following two facts:

  1. If $b > 1$, then $b^w$ is an increasing function of $w$; and,
  2. If $b > 1$, then $b^n - 1\geq n (b - 1)$ for all positive integers, $n$.

Let $b > 1$, $P = \{b^n : n \in \mathbb{N}\}$, and suppose (towards a contradiction) that the set, $P$, has an upper bound, $C$. Since $$C \geq b^n > n (b - 1),$$ it follows that $C/(b - 1) > n$ for all positive integers, $n$. This means that $\mathbb{N}$ is bounded above which contradicts the Archimedean property. Thus, $P$ does not have an upper bound.

Let $y > 0$, and $A = \{w\in \mathbb{R}: b^w < y\}$. Since $y$ is not an upper bound for $P$, there must be some positive integer, $N$, such that $b^N > y$. If $w$ is real, and $w \geq N$, then $b^w \geq b^N > y$, so $w \notin A$. The contrapositive of the previous statement tell us that if $w \in A$, then we must have $w < N$, and $N$ is an upper bound for $A$.

The Archimedean property guarantees that we have a positive integer, $M$, such that $y(b - 1) > 1/M$. It follows that $$b^{-M} < \frac{1}{M(b - 1)} < y,$$ and that $-M \in A$. Thus, $A$ is nonempty.

I think the problem with your approach is that you tried to prove $A$ was nonempty by looking for a root of $b$ that was less than $y$, but this is not possible if $y \leq 1$ because $b^{1/n} > 1$ for every positive integer, $n$. Instead, you have to look for negative integers powers of $b$ as shown above, and there are no cases to separate the proof into.

Can you take it the rest of the way? If $x = \sup A$, show that $b^x < y$ and $b^x > y$ are false statements from which we can conclude that $b^x = y$.

shoteyes
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Bernoulli's Inequality can be proven for integer exponents using induction, and extended to rational exponents also using induction. We will assume that for $b\gt0$ and all $x\in\mathbb{R}$, $b^x\gt0$.

Lemma: If $\boldsymbol{b\gt1}$ and $\boldsymbol{x\gt0}$, then $\boldsymbol{b^x\ge1+x\frac{b-1}b\gt1}$ $$ \begin{align} b^x &=\left(1-\frac{b-1}b\right)^{-x}\tag{1a}\\ &\ge1+x\frac{b-1}b\tag{1b}\\[6pt] &\gt1\tag{1c} \end{align} $$ Explanation:
$\text{(1a)}$: rewrite $b^x=(1/b)^{-x}$
$\text{(1b)}$: Bernoulli's Inequality for negative exponents $\left(-\frac{b-1}b\gt-1\right)$
$\text{(1c)}$: $x\gt0,b\gt1\implies x\frac{b-1}b\gt0$

Corollary: If $\boldsymbol{b\gt1}$, then $\boldsymbol{b^x}$ is Increasing

If $x\gt y$, then $$ \begin{align} b^x &=b^{x-y}b^y\tag{2a}\\[9pt] &\gt b^y\tag{2b} \end{align} $$ Explanation:
$\text{(2a)}$: property of exponents
$\text{(2b)}$: since $x-y\gt0$, $\text{(1c)}$ says that $b^{x-y}\gt1$

Upper Bound

For $b\gt1$ and $y\gt1$, if $x\gt b\frac{y-1}{b-1}$, then $$ \begin{align} b^x &\ge1+\color{#C00}{x}\frac{b-1}b\tag{3a}\\ &\ge1+\color{#C00}{b\frac{y-1}{b-1}}\frac{b-1}b\tag{3b}\\[3pt] &=y\tag{3c} \end{align} $$ Explanation:
$\text{(3a)}$: Lemma
$\text{(3b)}$: assumption
$\text{(3c)}$: simplify

For $b\gt1$ and $y\le1$, if $x\gt0$, then $b^x\gt b^0=1\ge y$.

Thus, if $b^x\lt y$, then $x\lt\max\!\left(b\frac{y-1}{b-1},0\right)$.

Therefore, $A=\{x\in\mathbb{R}:b^x\lt y\}$ has an upper bound of $\max\!\left(b\frac{y-1}{b-1},0\right)$.

Existence

For $b\gt1$ and $y\lt1$, if $x\lt\frac by\frac{y-1}{b-1}$, then $$ \begin{align} b^{-x} &\ge1\color{#C00}{-x}\frac{b-1}b\tag{4a}\\[3pt] &\gt1\color{#C00}{-\frac by\frac{y-1}{b-1}}\frac{b-1}b\tag{4b}\\ &=\frac1y\tag{4c} \end{align} $$ Explanation:
$\text{(4a)}$: Lemma $(-x\gt0)$
$\text{(4b)}$: assumption
$\text{(4c)}$: simplify

Thus, if $y\lt1$ and $x\lt\frac by\frac{y-1}{b-1}$, then $b^x\lt y$.

For $b\gt1$ and $y\ge1$, if $x\lt0$, then $b^x\lt b^0=1\le y$.

if $x\lt\min\!\left(\frac by\frac{y-1}{b-1},0\right)$, then $b^x\lt y$.

Therefore, $\left(-\infty,\min\!\left(\frac by\frac{y-1}{b-1},0\right)\right)\subset A=\{x\in\mathbb{R}:b^x\lt y\}$.

robjohn
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