Finding and proving supremum and infimum Sorry to bother you with such simple problems but I can't do it on my own, since my professor doesn't teach at all. I don't even know where to begin. 
I need to find the supremum and infimum of the following sets:


*

*$A=\{3n\mid n\in N\}$

*$B=\{2/n \mid n \in N\}$

*$A \cup B$

 A: First, note that $A$ is not bounded above. To see this, suppose that $k$ is an upper bound for $A$. Then, because of the definition of $A$, this means that $3n \leq k$ for all $n \in \mathbb N$. Equivalently, $n \leq k/3$. But this is absurd, because there is no largest natural number. So, $A$ is not bounded above and therefore it has no finite supremum. We can say that $\sup A = \infty$ in this case.
Next, we note that $n \geq 1$ for all $n \in \mathbb N$, so $3n \geq 3$ for all $n \in \mathbb N$. This means that $3$ is a lower bound for $A$. Moreover, there cannot be any greater lower bound for $A$, because $3 \in A$. So, $\inf A = 3$.
For $B$, first note that all of the elements of $B$ are positive, so $0$ is a lower bound for $B$. Moreover, if $x > 0$ then there is some $n$ for which $2/n < x$ (just take $n > 2/x$), so $x$ is not a lower bound for $B$. Therefore $0$ is the greatest lower bound: $\inf B = 0$.
Finally, note that since $n \geq 1$ for all $n \in \mathbb N$, we have $1/n \leq 1$ and so $2/n \leq 2$. Therefore, $2$ is an upper bound for $B$. Moreover, there cannot be any smaller upper bound, because $2 \in B$ (take $n = 1$). So $\sup B = 2$.
For $A \cup B$, note that in general, if we add elements to a set, its infimum must stay the same or decrease, and its supremum must stay the same or increase. Since $A \cup B$ contains both $A$ and $B$, its infimum cannot exceed the $\inf A$ or $\inf B$, and its supremum cannot be smaller than $\sup A$ or $\sup B$. We can express this as $\inf (A \cup B) \leq \min\{\inf(A), \inf(B)\}$ and $\sup(A \cup B) \geq \max\{\sup(A), \sup(B)\}$
Since $\min\{\inf(A), \inf(B)\} = \min\{3, 0\} = 0$ and $\max\{\sup(A), \sup(B)\} = \max\{\infty, 2\} = \infty$, this means that $\inf(A \cup B) \leq 0$ and $\sup(A \cup B) \geq \infty$. So right away this means that $\sup(A \cup B) = \infty$. What about $\inf(A \cup B)$? Can it be smaller than zero? This is only possible if $A \cup B$ contains some element smaller than zero, but it does not. Therefore, $\inf(A \cup B) \geq 0$. Combining this with $\inf(A \cup B) \leq 0$ deduced above, we conclude that $\inf(A \cup B) = 0$.
