[College Calc 01-01. The Reals] #02

We must note that each $A_{n+1}$ is indeed finite, as there are at most $|A_n|^2$ choices of the pair $(x,y).$

Each subset of a finite set is finite and thus has a maximum -- this is, in fact, the supremum of the subset, as you should check.

Hence we know that each $A_n$ must have the l.u.b. property.

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As for $A,$ however, things are different. $A$ is an infinite set. We can easily see it by noting that $\dfrac{1}{2^{n-1}} \in A_n \subset A$ for each $n.$

Sadly enough, it turns out that $A$ does not have the l.u.b. property. The reasoning is as follows:

$\dfrac{1}{\sqrt{2}}$ clearly does not belong to the set, since if it did, it must for some $A_n$ as well, which by definition only contains rationals.

Notice that $\dfrac{k}{2^{n-1}}\in A_n$ for each $n$ and $0\leq k\leq 2^{n-1}.$ We can use this as a sort of "grid" to keep inching towards $\dfrac{1}{\sqrt{2}}.$

Let $a_n$ be the largest integer $k$ satisfying

$\frac{k}{2^{n-1}}<\frac{1}{\sqrt{2}}.$

Note that this implies the following(can you explain why?):

$0< \frac{1}{\sqrt{2}} - \frac{a_n}{2^{n-1}} < \frac{1}{2^{n-1}}.$

From squeeze theorem , all of the expressions converge to zero, and thus we have

$\lim_{n\to\infty} \frac{a_n}{2^{n-1}} = \frac{1}{\sqrt{2}}.$

Letting

$K = \left\{\left.\frac{a_n}{2^{n-1}}\,\right|\, n\in\mathbb{Z}_+\right\}$

we immediately see that $K$ is bounded by $1\in A,$ and

$\sup K = \frac{1}{\sqrt{2}} \notin A$

which proves $A$ does not have the l.u.b. property.