proving existence of supremum in $\mathbb Q$

Question

I'm taking an analysis class and I'm a little confused about this question. Also I'm mostly a computer science guy so I'm not great at proof based math so I apologize if this is ignorant

Let $A = \{x : x \in \mathbb Q,\ x^3 < 2\}$

Prove that $\sup A$ exists. Guess the value of $\sup A$.

So from what I understand, if you're work in the set of rationals, you can't set a least upper bound but you can find the sup which in this case would be $2^{1/3}$ However, I'm not really sure how I could prove that. Any advice would be great. Thanks

Answer 1

If $A$ is any nonempty set of real numbers that has an upper bound, then $\sup A$ exists, by the completeness property of the real numbers. So it suffices to show that your set $A$ is nonempty and has an upper bound. I trust that you can show $A$ is nonempty (you just have to give an example of an element of $A$). To say that $A$ has an upper bound means that there is an element $r\in \mathbb{R}$ such that $x\leq r$ for all $x\in A$. That is, you want a number $r$ such that $x\leq r$ for all $x\in\mathbb{Q}$ such that $x^3<2$. Can you think of any such number?

(It turns out that $\sup A$ is indeed $2^{1/3}$, but you don't have to prove that to solve the problem--it just asks you to "guess" what you think $\sup A$ is!)

Answer 2

Observe that $A\neq \emptyset$ since $1 \in A$, and $A$ is bounded above by $2$, thus there exists $\sup(A)$. You show $\sup(A) = \sqrt[3]{2}$. This amounts to showing that for any small enough $\epsilon > 0$, there exists an $x \in A$ such that $x > \sqrt[3]{2} - \epsilon\iff (x+\epsilon)^3 > 2$. To this end, first choose an $n \in \mathbb{N}$ such that $\epsilon > \dfrac{1}{n}$, and let $x = 1+\dfrac{k}{n}$, you will solve for $k$ with $0 \leq k < n$ such that: $\left(x+\epsilon\right)^3 > 2$. Using Bernoulli inequality we have: $\left(x+\epsilon\right)^3 > \left(1+\left(\dfrac{k+1}{n}\right)\right)^3 > 1 + 3\left(\dfrac{k+1}{n}\right) > 2 \iff k > \dfrac{n-3}{n}$. Also you want that $k < n$. Observe such $k$ always exists because the interval $\left(\dfrac{n-3}{3}, n\right)$ has length $n - \dfrac{n-3}{3}=\dfrac{2n+3}{3} > 1$ which means it contains a natural number you call $k$, and you are done.