How do we know that $\sup$ exists for $\{w\in \Bbb R : b^w<y\}$

Question

This question is from Rudin's Principles of Mathematical Analysis:

This is exercise 7 of Ch 1:

Fix $b>1$ and $y>0$. Let $A$ be the set of all $w$ such that $b^w<y$ and show that $x=\sup A$ satisfies $b^x=y$.

I can't see why $\sup$ for $A$ exists.

Answer 1

To address your query about why $A$ is nonempty:

Note that by part a, you can choose $n$ so that $b^n$ is arbitrarily large, say $b^n>1/y$. [It suffices to pick $n>1/((b-1)y)$.] Then $b^{-n} < y$ since $b^{-n}b^n = 1$.

Answer 2

When can you take a supremum of a set $A$ of real numbers?

If you are ok with $+\infty$, all you need is non-emptiness of $A$.

If you want a finite number, $A$ needs to bounded as well.

Answer 3

Define $s$ to be any number such that $b^s \geq y$. Then, $s$ is an upperbound for your set, and by the least upper bound property of the real numbers the supremum of the set exists.