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Let $x\in (0,1)$ Compute with careful proof:

The greatest lower bound of $(x^n : n \in N)$ and the least upper bound of $(x^n : n \in N)$

Hint: For the infimum (greatest lower bound), first prove that if the greatest lower bound were strictly positive then there would be some $n\in N$ with $$x^n \lt \frac{\mbox{greatest lower bound}}{x}$$

Guess: Well I know the supremum of X^n is just X but having a hard time proving that is the case that there does not exist a smaller x which is not the least upper bound, while I can prove that X is an upper bound, proving it is the least upper bound is where I am having difficulty.

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Let $x\in (0,1)$. We continue with induction on $n$. First observe Since $0<x<1$ we have $0<x*x<1*x$ (i.e. $0<x^2<x$). Now suppose this is true for $n=k$. Then $0<x^k<x^{k-1}$. We must show that $0<x^{k+1}<x^k$. This is true since $0<(x^k)x<x^{k-1}x$ which is equivalent to $0<x^{k+1}<x^k$. Thus, $\forall n\geq 2$ we have $0<x^n<x^{n-1}$ assuming $0<x<1$.

Cool. So, suppose $\sup\{x^n: 0<x<1, n\in \mathbb{N}\}=x$. Since we have shown above $x>x^n$ $\forall n\geq 2$, we have that $x$ is the last element of the set, thus it is the least upper bound.

Now suppose $\inf\{x^n: 0<x<1, n\in \mathbb{N}\}=0$. Then we have $0<x^n$ $\forall n\in \mathbb{N}$. This is true as we proved it above. Now suppose for the sake of contradiction $\alpha$ is some other lower bound of the set of $x^n$, but $\alpha>0$.

Now you may use the hint to prove the rest.

Eoin
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