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I'm currently working through Baby Rudin and need help on exercise 1.7(f).

The question:

Let $A$ be the set of all $w$ such that $b^w < y$, and show that $x = \text{sup}(A)$ satisfies $b^x = y$.

Things given and proven in parts (a)-(e).

  1. $b > 1$, $y > 0$, $x \in \mathbb{R}$
  2. $\forall n \in \mathbb{Z}^+$, $b^n - 1 \geq n(b-1)$
  3. $b-1 \geq n(b^{1/n}-1)$
  4. $b^{1/n} < t$
  5. $b^w < y \implies b^{w+(1/n)} < y$, for sufficiently large $n$, and using the substitution $t = yb^{-w}$.
  6. $b^w > y \implies b^{w-(1/n)} > y$

I'm having trouble proving this. My approach so far is to try to prove that both $b^x < y$ and $b^x > y$ are false, so that $b^x = y$.

Proof (so far):

By definition, $\forall w \in A$, $x \geq w$. Assume $b^x < y$. $b^{x+1/n} < y \implies b^xb^{1/n} < y$ (This is where my proof falls apart, I'd like to use fact #4 to substitute in $t$, but it might invalidate my inequality. The same problem occurs if I assume $b^x > y$. Any ideas?)

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well you certainly have $$ b^x \le y $$ this suggests an approach based on

suppose $b^x \lt y$

can you use the facts you cite to show that $$ \exists z \gt x\\ b^z \lt y $$ a contradiction?

added: i see you have already got this idea, so maybe just tinker around a bit longer! property (5) looks promising.

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  • $\begingroup$ Thanks, z = x + 1/n is what I needed. $\endgroup$ – user114464 Dec 9 '13 at 5:12
  • $\begingroup$ good! in maths we often find ourselves caught between not seeing the wood for the trees and not seeing the trees for the wood. well done anyhow. there is nothing quite like the satisfaction of solving a puzzle that has had one going around in ever-decreasing circles. unfortunately for me that is a rare, though blissful, experience. $\endgroup$ – David Holden Dec 9 '13 at 7:44

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