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

This is exercise 7 of Ch 1: enter image description here

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.

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    $\begingroup$ Hint: look at condition (a) and add $1$ to both sides. Now find an $n$ such that $n(b-1)+1\gt y$... $\endgroup$ – Steven Stadnicki Oct 4 '17 at 17:12
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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 $b^{-n}b^n = 1$.

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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.

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  • $\begingroup$ I know that! What I meant that I can't find an upper bound and an element in set $A$. For example, I can see that $0\in A$ if $y>1$, but how to show that $A$ is nonempty if $0<y<1$? $\endgroup$ – Silent Oct 4 '17 at 17:09
  • $\begingroup$ Where exactly did you get stuck? Rudin has a lot of impossible problems,but this isn't one of them-he gives you a clear path to follow on this. $\endgroup$ – Mathemagician1234 Oct 4 '17 at 17:12
  • $\begingroup$ @Mathemagician1234, will you just please prove that $A$ is nonempty? That will be a big help! I meam, it seems very difficult to find a $w$ if I take $b$ very large and $y$ very close to zero. $\endgroup$ – Silent Oct 4 '17 at 17:18
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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.

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