[College Calc 01-01. The Reals] #03

A

We can easily check that $\sup \left\{\left.-\dfrac{1}{\sqrt{n}}\,\right|\, n\in \mathbb{Z}_+\right\} = 0\notin A.$ Thus $A$ does not have the l.u.b. property.

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B

As for $B,$ suppose it does not satisfy the l.u.b. property. Let $X$ be a subset of $B$ such that

$\sup X = k\notin B.$

Clearly $0< k\leq 1$ which means that if $n>m^2$ and $\dfrac{m}{n}>\dfrac{k}{2}$ then $\dfrac{1}{\sqrt{n}}=\dfrac{\sqrt{n}}{n}>\dfrac{m}{n}>\dfrac{k}{2}$ such that

$0<n<\frac{4}{k^2}.$

Notice, however, that the number of such $\dfrac{m}{n}$ is finite, as the choice of $n$ is now finite. This means there exists a maximal element $c$ of $C$ such that

$c<k$

which is a contradiction, because $c$ itself would be the l.u.b of $X$ . Therefore, $B$ has the l.u.b. property

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C

Again, suppose it does not satisfy the l.u.b. property, and let $X$ be a subset of $C$ such that

$\sup X = k\notin C.$

Clearly $0.888\cdots = \dfrac{8}{9} < k < 1 = 0.999\cdots.$

In the decimal expansion of $k,$ locate the digit $d$ that is neither 8 nor 9, and is closest to the decimal point. (Of course, excluding the starting 0.) Suppose $d$ is at the $n$ -th decimal place. If it is zero, the consecutive zeros after that point must have been finite.*

Then you will notice that the largest $\overline{0.x_1x_2x_3\cdots} < k$ can ever be is $k'=\overline{0.x_1x_2\cdots (x_{n-1}-1) 999\cdots}.$ It can never be larger.

Differently said, each $x\in X\subset C$ , since it satisfies $x < k,$ must also satisfy $x \leq k'.$ This is a contradiction, because $\sup X = k'\neq k$ . Therefore $C$ does indeed have the l.u.b. property.

(*This finite consecutive zeros condition is due to the fact that if the consecutive zeros were infinite, we can subtract $1$ from the previous digit and replace all zeros with $9$ without changing the value of $k,$ which then means that $k\in C.$ )