Consider the set $\{z \in \mathbb{R}_+ : z^p < x \}$ where $p \in \mathbb{N}$.

I want to show that the least upper bound $z_0$ of this set satisfies $z_0^p = x$.

That the least upper bound exists is mandated by the completeness of the reals (right?).

  1. Then, suppose $z_0^p < x$, then by density of rationals, there exists some $\epsilon > 0$ so that $(z_0 + \epsilon)^p < x$ hence contradiction.
  2. Suppose $z_0^p > x$, then argue similarly.

I am a beginner in analysis so I want to make points 1 and 2 more mathematically rigorous / correct. Any help would be appreciated.

  • $\begingroup$ The "argue similarly" is a bit shady. You can find a full proof Rudin's Principles of Mathematical Analysis, Theorem 1.21 in page 19. $\endgroup$ – Reveillark Sep 27 '19 at 2:11
  • $\begingroup$ My guess is that he's doing the Rudin exercise which asks you to prove that we can define $x^{1/p}$ for all $p \in \mathbb{N}$ and all $x \in \mathbb{R}_+$. $\endgroup$ – Charles Hudgins Sep 27 '19 at 2:15
  • $\begingroup$ Assuming that $z_0^p < x,$ the density of the rationals would imply there is a rational number $r$ such that $z_0^p < r < x.$ How does that imply there is an $\epsilon$ with the property you propose? (especially when learning a topic, you ought to avoid these leaps of logic) $\endgroup$ – Brian Moehring Sep 27 '19 at 2:15
  • $\begingroup$ @BrianMoehring How do I contradict the first point then? intuitively it is clear that there is a contradiction. f(.) = (.)^p is an increasing function also. $\endgroup$ – ironX Sep 27 '19 at 2:42
  • $\begingroup$ Well, do you know how to prove that $f(x) = x^p$ is continuous? Proving such an $\epsilon$ exists is basically the same thing. $\endgroup$ – Brian Moehring Sep 27 '19 at 2:58

Let $A=\{z>0:z^p<x\}.$ Let $B=\{z>0:z^p>x\}.$ Let $A^*=\{xz^{1-p}: z\in A\}.$

$(***)$. If $a,b$ are positive then $a<b \iff a^p<b^p.$

Use $(***)$ to show that $A^*=B,$ and that every $a\in A$ is less than every $b \in B.$

Show that $A \ne \emptyset,$ and hence also $B=A^*\ne \emptyset.$

Now neither of $A, B$ is empty and any $a\in A$ is a positive lower bound for $B$ while any $b\in B$ is an upper bound for $A,$ so the values $z_0=\sup A$ and $y_0=\inf B$ exist.

We have $z_0\le y_0.$ Because whenever $A,B$ are non-empty subsets of $\Bbb R$ such that every $a\in A$ is less than every $b\in B ,$ we have $\sup A\le \inf B.$

Suppose by contradiction that $z_0<y_0.$ By $(***)$ there is at most one $w\in \Bbb R_+$ such that $w^p=x.$ That is, there is at most one $w\in \Bbb R_+$ such that $w\not \in A\cup B.$ Now if $z_0<y_0$ then there are infinitely many $w$ such that $z_0<w<y_0$ and hence some $w\in (z_0,y_0)\cap (A\cup B)$. But such a $w$ would either be a member of $A$ that's greater than $z_0=\sup A$ (which is absurd), or a member of $B$ that's less than $y_0=\inf B$ (also absurd). So by contradiction we deduce $z_0=y_0.$

Let $U$ be the set of upper bounds for $A.$ Let $V$ be the set of positive lower bounds for $B.$ Use $A^*=B$ and $(***)$ to show that $V=\{xu^{1-p}:u\in U\} $. And hence $$z_0=y_0=\max V=\max \{xu^{1-p}:u\in U\}=\frac {x}{(\min U)^{p-1}}=\frac {x}{z_0^{p-1}}.$$

Addendum. When we reach $z_0=y_0$ in the proof above, we can finish differently as follows: Show that $\max A$ does not exist. Hence by $B=A^*$ and $(***)$ we infer that $\min B$ does not exist either. So $z_0=\sup A \not \in A$ and $z_0=y_0=\inf B\not \in B.$ So $0<z_0\not \in A\cup B,$ and by definition of $A$ and $B$, this implies that $z_0^p=x.$

To show that $\max A$ does not exist, take $a \in A$ and $0<d<a.$ Then for integer $j$ with $1\le j\le p$ we have $a^{p-j}d^j\le a^{p-1}d.$ So by the Binomial Theorem we have $$(a+d)^p=a^p+\sum_{j=1}^p\binom {p}{j}a^{p-j}d^j\le $$ $$\le a^p +\sum_{j=1}^p\binom {p}{j}a^{p-1}d=a^p+(2^p-1)a^{p-1}d.$$ So if $d\in (0,a)$ and if $d<\frac {x-a^p}{(2^p-1)a^{p-1}}$ then $(a+d)^p<x,$ so $a<a+d\in A.$

  • $\begingroup$ An easy way to show that if $z<y$ then there are infinitely many $w$ such that $z<w<y$ is that for each $n\in \Bbb N$ we have $z<\frac {(n+1)z+y}{n+2}<\frac {nz+y}{n+1}<y.$ $\endgroup$ – DanielWainfleet Sep 27 '19 at 5:31

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