prove the set is bounded above - solution check (with progress and solution Let $a\in \Bbb R, a>0$ and $p\in$ $\Bbb N\backslash\{0\}$ and $B:=\{x\in \Bbb R: 0 \le x, x^p\le a\}$

*

*prove that B is bounded above

*For $ c:= \sup B$; $c^p=a$

*The equation $y^p=a$ in $(0, \infty)$ has the unique solution of c.

So my attempt:
Since $a> 0 $ and $p\ge1$ and $\forall x \in B,x\ge 0$, $x^p\ge0$, $(x+1)^p>x^p$. And since $x\in\Bbb B$ is a real number, $(x+1)$ must also be a real number. Hence the set B is bounded above. (However, I do believe that I should mention something about square root for real numbers, that's the title of the question) no sure thou.
For b) Since we proved the set is bounded above, there must exist a supremum. A supremum is an upper bound itself. Since $x^p\le a,x\le\sqrt[p]{a}$. And also since $(x+\ell)^p>x^p \forall\ell>0$, we can conclude the supremum of b is $x^p=a=c$
How does this look to you? I won't show my c since I think a and b are not very correct...
 A: It is more or less correct, although you have some unnescessary and some missing arguments. If I’m correct this is part of the proof to the existence of the $p$-th root, thus you should not use roots here.
For 1: Rather than $(x+1)^p$ consider $(x+t)$ for any $t>0$. The binomial formula gives you $x^p+R$ with $R>0$. Thus the function is strictly monotonous. What you are now missing is a good argument for why this means that there exists such a bound. For example you can argue that $x^p\xrightarrow{x\to\infty} \infty$. Thus there exists some $x_0$ so that $x_0^p > a$. Due to monotony the set is thus bounded by $x_0$.
For 2: You do not need the roots. If $c$ is the sup then there exists a sequence $x_n$ in $B$ that converges to $c$. But as $x_n^p\leq a$ we also have for the limit $c^p\leq a$. But if $c^p< a$ then you need to argue that there is some $y>c$ with $y^p\leq a$. In that case: It is easy to see that $c\neq 0$, as there exists an $n\in\mathbb N$ so that $1/n < \min(a,1)$. Then $(1/n)^p\leq 1/n) < a$. But if $c\neq 0$ then we can take $1\leq a/c^p$. Now if this was $<$ then take $0<a/c^p-1$.
From the binomial formula we get that $(1+x)^p = \sum_{i=0}^p {p\choose i} x^i < 1+C_px$ for $0\leq x\leq 1$ for some constant $C_p\in\mathbb N$. (Since all $x^i \leq x^1$ for $i\geq 1$, then $C_p = \sum_{i=1}^p {p\choose i} = 2^p - 1$).
Thus we find an $n$ so that $1/n < a/(c^pC_p)$ and thus $(1+1/n)^p < 1+C_p/n < a/c^p$. Then $c(1+1/n)>c$ but $(c(1+1/n))^p < c^p a/c^p = a$.
For 3 you’ll need to show that this solution $c$ is unique, so if $y_1^p=y_2^p = a$, then $y_1=y_2$. For this you can use the strict monotonity.
