# least upper bound property of reals

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.

• 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. – Reveillark Sep 27 '19 at 2:11
• 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 $x \in \mathbb{R}_+$. – Charles Hudgins Sep 27 '19 at 2:15
• 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) – Brian Moehring Sep 27 '19 at 2:15
• @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. – ironX Sep 27 '19 at 2:42
• Well, do you know how to prove that $f(x) = x^p$ is continuous? Proving such an $\epsilon$ exists is basically the same thing. – Brian Moehring Sep 27 '19 at 2:58

## 1 Answer

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

$$(***)$$. If $$a,b$$ are positive then $$a

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 $$z_0 By $$(***)$$ there is at most one $$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 then there are infinitely many $$w$$ such that $$z_0 and hence some $$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 $$0 and by definition of $$A$$ and $$B$$, this implies that $$z_0^p=x.$$

To show that $$\max A$$ does not exist, take $$a \in A$$ and $$0 Then for integer $$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 so $$a

• 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 $z<\frac {(n+1)z+y}{n+2}<\frac {nz+y}{n+1}<y.$ – DanielWainfleet Sep 27 '19 at 5:31