Our intuition shows us that $\sup A = \inf A = 1$, and one way to demonstrate this is to show that $A = \{1\}$. However, we can proceed in a different manner.
We will proceed in two steps:
- First, we will show that $1$ is an upper bound for $A$.
- Then we will assume that $\sup A < 1$ and derive a contradiction.
To show that $1$ is an upper bound, we must have for every $x \in A$, $x \leq 1$. To this end, let $x \in A$. By definition, we have
$$\text{For EVERY } n \in \mathbb{N}, x \in \Big(1-\frac{1}{n}, 1+\frac{1}{n}\Big).$$
This is important, because we will show that, in fact, $x$ does not belong to every interval of that form as follows.
Since $x \in \Big(1-\frac{1}{n}, 1+\frac{1}{n}\Big)$ it follows that either $x \leq 1$ or $1 < x$. If $x \leq 1$, then we are done, because we want to show that $1$ is an upper bound for $A$.
Now suppose that $1 < x$. Then $0 < x-1$. By the Archimedean Principle, there exists a positive integer $M$ such that
$$\frac{1}{M} < x-1,$$
which implies that
$$1 + \frac{1}{M} < x.$$
But $1-\frac{1}{M} < 1+\frac{1}{M} < x$, which means that
$$x \notin \Big(1-\frac{1}{M}, 1+\frac{1}{M}\Big),$$
which is a contradiction, because $x$ belongs to every interval of the form $\big(1-\frac{1}{n}, 1+\frac{1}{n}\big)$.
Consequently, we must have $x \leq 1$ for every $x \in A$, and thus $1$ is an upper bound for $A$.
Since $A$ is a nonempty set bounded above, the Completeness axiom asserts that $\sup A$ exists. For sake of contradiction, suppose that $\sup A < 1$.
What we need to do is prove that under the assumption $\sup A < 1$, we can show that $\sup A$ is not an upper bound for $A$. But if $\sup A$ is not an upper bound for $A$, then we should be able to find an $x \in A$ such that $\sup A < x$.
And if such an $x \in A$ exists, then we must have $x \in \big(1-\frac{1}{n}, 1+\frac{1}{n}\Big)$ for every positive integer $n$.
Can you think of a number $x$ that belongs to every interval $\big(1-\frac{1}{n}, 1+\frac{1}{n}\big)$ such that $\sup A < x$? If you can, continue reading by hovering your mouse over the yellowish area below.
The $x$ we seek is $1$. Since for every positive integer $n$
$$1-\frac{1}{n} < 1 < 1+\frac{1}{n}$$
it follows that $1 \in A$.
Since $\sup A$ is an upper bound for $A$, we must have $1 < \sup A$.
But this contradicts our assumption that $\sup A < 1$.
Therefore, $1 \leq \sup A$. Because $1$ is an upper bound for $A$, we conclude that $\sup A = 1$.
If you understand the proof above, it is possible to clean it up a bit.
To show that $\inf A = 1$, we will show that $1$ is a lower bound for $A$. To this end, let $x \in A$, so that $x \in \big(1-\frac{1}{n}, 1+\frac{1}{n}\big)$ for every positive integer $n$. Suppose that $x < 1$. Then $0 < 1-x$. By the Archimedean Principle, there exists a positive integer $m$ such that
$$\frac{1}{m} < 1- x \implies x < 1 - \frac{1}{m}$$
But since
$$1 - \frac{1}{m} < 1+\frac{1}{m},$$
it follows that
$$x \notin \Big(1-\frac{1}{m}, 1+\frac{1}{m}\Big),$$
contradicting the assumption that $x$ belongs to every interval of the form $\Big(1-\frac{1}{n}, 1+\frac{1}{n}\Big)$. Therefore, $1 \leq x$ for all $x \in A$, so $1$ is a lower bound for $A$.
Since $A$ is bounded below and nonempty, $\inf A$ exists. Suppose $1 < \inf A$. Then since $1 \in A$ and $\inf A$ is a lower bound for $A$, we must have $\inf A \leq 1$, a contradiction. Therefore, $\inf A = 1$.
We have proven that $\sup A = \inf A = 1$.
Now onto the set $B$.
To show that $\sup B = 2$, we proceed as we did in the preceding proof: First, we demonstrate that $2$ is an upper bound for $B$, then we prove that $2$ is the smallest upper bound for $B$.
For every $x \in B$, we have $x^3 < 8$. Consequently, $x < 2$, so $2$ is an upper bound for $B$. Since $B$ is bounded above, $\sup B$ exists. Suppose that $\sup B < 2$. Then by the Archimedean Principle, there exists positive integer $k$ such that
$$\frac{1}{k} < 2 - \sup B,$$
and thus
$$\sup B + \frac{1}{k} < 2.$$
But this implies that
$$\Big(\sup B + \frac{1}{k}\Big)^3 < 8,$$
so $\sup B + \frac{1}{k} \in B$.
Since $\sup B$ is an upper bound for $B$ and $\sup B + \frac{1}{k} \in B$, it follows that
$$\sup B + \frac{1}{k} \leq \sup B,$$
which is a contradiction, because $1/k > 0$. We conclude that $\sup B = 2.$
To show that $B$ is not bounded below, we proceed by contradiction. Assume to the contrary that $B$ is bounded below by $L$. We state a couple observations
- Since $L \leq x$ for all $x \in B$, we have $L^3 \leq x^3$.
- Since $0 \in B$ and $L$ is a lower bound for $B$, we must have $L \leq 0$. Similarly, $L < 1$. Therefore
$$L - L^2 = L(1-L) \leq 0.$$
We will now demonstrate that $L$ is not a lower bound for $B$ by showing that $L-1 \in B$. Obviously, $L-1 < L$, but we see from the two observations above that for every $x \in B$:
\begin{align}
\big(L-1\big)^3 &= L^3 - 3L^2 + 3L - 1\\[5pt]
&= L^3 + 3L\big(1-L\big) - 1\\[5pt]
&\leq L^3 + 0 - 1\\[5pt]
&< L^3 \\[5pt]
&\leq x^3 \\[5pt]
&< 8
\end{align}
Therefore, $\big(L-1\big)^3 < 8$, so $L-1 \in B$. Because $L$ is a lower bound for $B$, we must have $L \leq L-1$, which is a contradiction.
Therefore, there are no lower bounds for $B$, and thus $B$ is unbounded below.
Your final paragraph outlines a perfectly reasonable strategy for (a). So let's do that. And remember the defining property of $\sup$ (and analogously for $\inf$): It is the least upper bound, meaning all other upper bounds are larger. So any inequality of the form $\sup(X) \leq t$ is best proven by showing that $t$ is an upper bound for $X$.
- $\sup(A\cup B)\leq \sup\{\sup(A), \sup(B)\}$: Let $x = \sup\{\sup(A), \sup(B)\}$, and take an element $a\in A\cup B$. Either $a\in A$, which means $a\leq \sup(A)\leq x$, or $a\in B$, which means $a\leq \sup(B)\leq x$. Either way we get $a\leq x$. This means $x$ is an upper bound for $A\cup B$.
- $\sup(A\cup B)\geq \sup\{\sup(A), \sup(B)\}$: Clearly, by using (b) ("larger set means larger $\sup$"), we have $\sup(A\cup B)\geq \sup(A)$, and just as clearly $\sup(A\cup B) \geq \sup(B)$. Thus $\sup(A\cup B)$ is an upper bound for $\{\sup(A), \sup(B)\}$.
The $\inf$ proof is entirely analogous, except we flip all the inequality signs, and change "upper" into "lower" (same goes for (b) and (c) as well).
Now for (b). This time there is only one inequality. It's rather simple to prove, though, using the defining property: We have that $\sup(B)$ is an upper bound for $B$, so it must be an upper bound for $A$.
The proof for (c) is very similar to half the proof for (a): again by using (b), clearly $\sup(A\cap B)\leq \sup(A)$, and just as clearly, $\sup(A\cap B)\leq \sup(B)$. Thus $\sup(A\cap B)$ is a lower bound for $\{\sup(A), \sup(B)\}$. (One could go the other way too, showing that $\inf\{\sup(A), \sup(B)\}$ is an upper bound for $A\cap B$. That would be more akin to the other half of the proof of (a). This time there isn't equality, because the two halves show the same inequality, rather than opposite inequalities the way they did in (a).)
Finally, we have (d). They give one hint as to what you could look at, but I like to do it simpler. Let us say our partially ordered set has three elements, two are incomparable, and the third is larger than both (the power set on a set of two elements yields a similar example). Then let $A$ and $B$ each consist of one of the two incomparable elements. Then the set $\{\sup(A), \sup(B)\}$ has no maximum as its elements are incomparable, by design.
Best Answer
Your reasoning for (a) is incorrect: if you fix $m=1$ and let $n \rightarrow \infty,$ we get $1/n \rightarrow 0.$ So $\inf A = 0.$
The others seem fine. Usually in analysis $\mathbb N$ does not include $0$ and judging by (b) I would assume this convention is taken here. It is likely that the author lists his notational conventions near the start of the book, so you should check there.