I've been attempting to prove the statement below, but I am stuck at one particular step of my proof and would appreciate any hint or clue on how I can move forward.
Statement:
Given two numbers $a$ and $n$, where $a > 0$ is real and $ n\geqslant 2$ is an integer.
Consider $S = \{ x \in \mathbb{R} \ | \ x \geqslant 0 \ \ \text{and} \ \ x^n < a \} $.
Then $\alpha = \sup S$ exists in $\mathbb{R}$ and moreover $\alpha^n = a$.
Proof attempt:
Since $0 \in S$ and $x \leqslant \max (a,1)$ for all $x \in S$, the completeness property of reals furnishes $\alpha = \sup S$.
To show $\alpha^n = a$ we assume the opposite.
I managed to show $\alpha^n < a$ produces a contradiction. I did this by considering $(\alpha + \epsilon)^n$ where $0 < \epsilon < 1$ and showed that $(\alpha + \epsilon)^n < a$. This implies $\alpha + \epsilon \in S$ and that contradicts the supremum property of $\alpha$.
However I'm unable to progress with the case $\alpha^n > a$. Here's how far I get:
The strategy is to find a number smaller than $\alpha$ that bounds $S$ from above; then we will have reached a contradiction.
Take $0 < \epsilon < \alpha$ and consider
$ \displaystyle (\alpha - \epsilon)^n = \alpha^n - n\alpha^{n-1} \epsilon + \sum_{k=2}^n {n \choose k}\alpha^{n-k}\epsilon^k(-1)^k > \alpha^n - n\alpha^{n-1}\epsilon - \sum_{k=3,5,7,...}^n {n \choose k}\alpha^{n-k}\epsilon^k $
From here I am not able to progress any further. I have a hunch that I should be able to show $(\alpha - \epsilon)^n > a$, which would settle the case.
In the last step I removed all positive terms with even $k$.
What would be desirable now is to rid ourselves of all higher powers of $\epsilon$, so that I finally can choose $\epsilon = C(\alpha^n - a)$ for some judiciously chosen constant $C$ that does not violate the restriction $0 < \epsilon < \alpha$.
If anybody has any hints/clues to share, please do. :)
I feel I'm so close to finishing this one but that darn sum seems impossible to deal with...
Thanks!
You want that if $\alpha^n>a$ there is $\beta$ with $0<\beta<\alpha$ and $\beta^n>a$. Try $\beta=\alpha(1-t)$ where $0<t<1$ is to be determined. By Bernoulli's inequality, $(1-t)^n\ge 1-nt$ and so $\beta^n\ge\alpha^n(1-nt)$. If $\alpha^n>a$ we can find a small positive $t$ such that $\alpha^n(1-nt)>a$.