Bounded above/below for set A Consider the set of real numbers:
$A = \{ \sqrt[n]{n}| n \in \mathbb{N}_{>0} \}$ and $A \subset \mathbb{R}$.
Prove that A is bounded above and below in $\mathbb{R}$.
Note: we are instructed to use what we proved in another question: $(n+1)^n < n^{n+1}$.
Edits: I forgot to note that $n \geqq 3$ is necessary for the above inequality. Thanks for the correction.

My thoughts:
I was thinking of proving by contradiction, as we had always been doing in class for supremum and infimum.
So, assume there is not an upper bond, meaning there is no $m \in \mathbb{R}$ such that m > a for all $a \in \mathbb{R}$.
I picked $m = \sqrt[n+1]{n+1}$ (which I am worried is very wrong.
Then, by the inequality we have &  m > a = $\sqrt[n]{n}$, I am able to prove by contradiction -- while this feels very wrong to me generally.
Any suggestion on what I should do?
Thanks in advance.
 A: Clearly the set is bounded from below, as its elements are positive.
And yes, your proof is wrong. Firsly, it seems like you tried to unessecarily prove by contradiction. Your proof by contradiction assumes the opposite of your goal, and then proves your goal saying that is a contradiction, while you could have just reached your goal without going through assuming its opposite...
Secondly, the upper bound can not depend on the elements of the set. The upper bound must be a certain velue, independant of $n$, in this case.
It is often more simple to just prove that there exists an upper bound, by finding it! Here I'll prove $10$ (or better, $\sqrt[3]{3}$...) is an upper bound using induction.
By the given inequality, for all $3\leq n\in \mathbb{N}$ (you forgot to note that the inequality is not true for $n=1, 2$),
$$(n+1)^n<n^{n+1} \implies (n+1)^{\frac{1}{n+1}}<n^{\frac{1}{n}}$$
All elements of $A$ are of the form $n^{\frac{1}{n}}$. For $n=1, 2, 3$, it is obvious that $10$ (for example) is an upper bound of the elements. Now, we will prove by induction that every other element is also smaller than $10$. We've seen it for $n=3$, so now we will assume $n^{\frac{1}{n}}<10$ (for some $3\leq n\in \mathbb{N}$), and show $(n+1)^{\frac{1}{n+1}}<10$. This is immediate using the above inequality:
$$(n+1)^{\frac{1}{n+1}}<n^{\frac{1}{n}}<10$$
So overall, for all $n\in \mathbb{N}$, $n^{\frac{1}{n}}<10$. In other words, all elements of $A$ are smaller than $10$, so by definition, $A$ is bounded from above.
A: The problem boils down to: prove that the sequence $(\sqrt[n]n)_{n\ge1}$ is bounded.
One could argue that it follows from the fact that it is convergent (its limit is $e^0=1,$ since $n^{1/n}=e^{\frac{\ln n}n}$ and $\lim_{n\to+\infty}\frac{\ln n}n=0$) but let us rather follow the instruction you received.
It is obviously bounded from below, e.g. by $0.$
Your inequality $(n+1)^n < n^{n+1}$ holds only for $n\ge3.$ Taking the $n(n+1)$-th root on both sides, it implies: $\forall n\ge3\quad(n+1)^{1/(n+1)}<n^{1/n},$ i.e. the subsequence $(\sqrt[n]n)_{n\ge3}$ is decreasing, whence
$$\forall n\ge3\quad\sqrt[n]n\le\sqrt[3]3$$
(whereas by the same reasoning, $(1+1)^1>1^{1+1}\Rightarrow\sqrt[2]2>\sqrt[1]1$ and $(2+1)^2>2^{2+1}\Rightarrow\sqrt[3]3>\sqrt[2]2$).
Therefore, $A$ has an upper bound, and even a greatest element:
$$\max(A)=\max(\sqrt[1]1,\sqrt[2]2,\sqrt[3]3)=\sqrt[3]3.$$
Incidentally, note that we also have $\min(A)=\sqrt[1]1=1.$
