We must prove the following implication: If \(z\in A\), then \(z\leq 3\).

 

We will instead prove the contrapositive: If \(z>3\), then \(z\notin A\).

Proof. We will prove the statement by contrapositive.

Suppose \(z>3\). By a theorem from class, since \(z>3\geq 0\) we see that \(z^2>3^2=9\). From this, we see that \(z\notin A\). \(\Box\)

Proof. Since \(a>0\) implies \(\frac{1}{a}>0\), we see that \(1-\frac{1}{a}<1\) for all \(a>0\). That is, \(1\) is an upper bound for the set \(\{1-\frac{1}{a} : a>0\}.\)

Let \(v<1\) be arbitrary. Note that \(1-v>0\), and hence

\[a_{0}:=\frac{2}{1-v}>0.\]

We see that \(1-\frac{1}{a_{0}}\in \{1-\frac{1}{a} : a>0\}\). Moreover,

\[1-\frac{1}{a_{0}} = 1-\frac{1}{\left(\frac{2}{1-v}\right)} = 1-\frac{1-v}{2} = \frac{1+v}{2}.\]

From our assumption that \(v<1\) we see that

\[\frac{1+v}{2}>\frac{v+v}{2} = v.\]

Thus, we see that \(1-\frac{1}{a_{0}}>v\), and hence \(v\) is not an upper bound for the set \(\{1-\frac{1}{a} : a>0\}\).

Thus, we have shown that \(1\) is an upper bound for the set, and if \(v<1\) then \(v\) is not an upper bound, hence \(1\) is the least upper bound. \(\Box\)

Proof. By a theorem from class we know that if \(a>0\), then \(\frac{1}{a}>0\). Thus \(0\) is a lower bound for the set \(\{\frac{1}{a}:a>0\}\).

Now, assume \(x>0\). We want to show that \(x\) is not a lower bound. By that same theorem, we know that \(\frac{1}{x}>0\), and hence \(\frac{2}{x}>0\) as well. From this we can see that

\[\frac{x}{2}= \frac{1}{\left(\frac{2}{x}\right)}\in\left\{\frac{1}{a} : a>0\right\}.\]

Since \(\frac{x}{2}<x\), this shows that \(x\) is not a lower bound for the set

\(\{\frac{1}{a} : a>0\}\). Therefore, \(\inf\{\frac{1}{a} : a>0\}=0\). \(\Box\)