Thread: Supremum Question for real analysis

1. Supremum Question for real analysis

This problem shows the existence of an nth root.
Let a > 0 and n >= 3 be an integer. Define the set S = {x > 0 : x^n <= a}.
1. Show that S is bounded above and thus that b = sup S exists.
2. Show that b^n = a. To do this, show that it cannot be true that b^n < a or b^n > a.

2. Originally Posted by flaming
This problem shows the existence of an nth root.
Let a > 0 and n >= 3 be an integer. Define the set S = {x > 0 : x^n <= a}.
1. Show that S is bounded above and thus that b = sup S exists.
2. Show that b^n = a. To do this, show that it cannot be true that b^n < a or b^n > a.
Clearly the $\sqrt[n]{a}$ is an upper bound for $S$ so $b \leqslant \sqrt[n]{a} \Rightarrow \quad b^n \leqslant a$.

If
$\begin{array}{rcl}
{b^n < a} & \Rightarrow & {b < \sqrt[n]{a}} \\
{} & \Rightarrow & {\left( {\exists x \in \left( {b,\sqrt[n]{a}} \right)} \right)\left[ {b^n < x^n < a} \right]} \\
{} & \Rightarrow & \mbox{contradiction} \\ \end{array}$

3. The problem is supposed to be showing that the nth root exists, so you can't use the nth root function to show that
the nth root function exists.

Otherwise it would be circular, that is, you would be assuming what you are trying to prove.

4. Originally Posted by flaming
This problem shows the existence of an nth root.
Let a > 0 and n >= 3 be an integer. Define the set S = {x > 0 : x^n <= a}.
1. Show that S is bounded above and thus that b = sup S exists.
2. Show that b^n = a. To do this, show that it cannot be true that b^n < a or b^n > a.
Let's do it the most elementary way since I presume this is the very beginning of your course.

1. For the existence of the supremum, it would be easier to define $S=\{x\geq 0\,:\,x^n\leq a\}$, so that it is obvious that $S$ is non-empty (it contains 0). Otherwise, this should be proven.
Anyway, to show that $S$ is bounded from above, you can notice that, if $x\geq a+1$, then, because $a+1\geq 1$, $x^n\geq (a+1)^n\geq (a+1)^2 =a^2+2a+1> 2a > a$ hence $x\notin S$, and $a+1$ is an upper bound for $S$.

2. Suppose by contradiction that $b^n. We must show that there exists $x>b$ such that $x^n\leq a$. To do this, I'll use the following computation:
if $0<\varepsilon<1$, then $(1+\varepsilon)^n-1=\varepsilon(1+(1+\varepsilon)+\cdots+(1+\varepsi lon)^{n-1})\leq \varepsilon (1+2+\cdots+2^{n-1})
\leq 2^n \varepsilon$
, hence $(1+\varepsilon)^n\leq 1+2^n\varepsilon.$

This is because of the usual formula $a^n-1=(a-1)(1+a+\cdots+a^{n-1})$ with $a=1+\varepsilon$.

Then, for $0<\varepsilon<1$, $(b(1+\varepsilon))^n=b^n(1+\varepsilon)^n\leq b^n+b^n 2^n \varepsilon$.
Let's choose $\varepsilon=\min\left(\frac{a-b^n}{b^n2^n},1\right)$. Then $0<\varepsilon<1$ and thus $(b(1+\varepsilon))^n\leq b^n+(a-b^n)=a$, so that $b(1+\varepsilon)\in S$. However, $b. This contradicts the fact that $b=\sup S$.

For the other part, this is the same idea, using $(1-\varepsilon)b$ with an $\varepsilon$ small enough so that $a<(1-\varepsilon)^n b^n$. You can use a similar estimate: for $0<\varepsilon<1$, we have
$(1-\varepsilon)^n-1 = -\varepsilon(1+(1-\varepsilon)+\cdots+(1-\varepsilon)^{n-1})\geq -n\varepsilon$, hence $(1-\varepsilon)^n\geq 1 - n\varepsilon$.