Let A and B be nonempty and upper bounded real positives sets. Define C... Let $A$ and $B$ be nonempty and upper bounded subsets of real positive numbers.
Define $C$ as:
$$C = \left\{\frac 1{a^{2}+b} : a \in A\wedge b\in B\right\}$$
Calculate the greatest lower bound of $C$.
What happens if $A$ or $B$ is not upperbounded?
 A: Since $A$ and $B$ are bounded above, we can consider the positive real numbers $L$ and $M$ where 
$$L = \sup(A), \quad M = \sup(B)$$
Now it may be the case the $L$ and $M$ are not actually elements of $A$ and $B$ respectively. However, this isn't really a concern since the question simply asks for the greatest lower bound of $C$, and this lower bound does not necessarily need to be an element of set $C$ itself. As an example of this idea, we see that the interval $(0,1]$ is bounded below by the number $0$ even if $0$ isn't actually in the interval itself. 
Now, observe that $C$ consists of positive fractions of the form 
$$\frac{1}{a^2 +b}$$
It is important to note that these fractions are positive since $a$ and $b$ are positive numbers. Now, we simply note the elementary detail that positive fractions become smaller in value as the denominator gets bigger and bigger. This means that we will find the smallest possible value of $\frac{1}{a^2 +b}$ when we find the largest value that the denominator may take on. 
So, what is the largest value that the denominator $a^2 + b$ may take on? Well, since $a \in A$ and $L = \sup(A)$, $L$ itself is the largest potential value that  $a$ may take. If it happens to be the case that $L \notin A$,then the numbers $a$ become arbitrarily close to $L$. We can use a similar argument to show that $M$ is the largest potential value that $b$ may take. Thus, the largest potential value that the denominator may be is simply $L^2 + M$. Therefore, 
$$\frac{1}{L^2 + M}$$
is the greatest lower bound of $C$ (i.e. no  fraction in $C$ can be smaller than this). 
If $A$ or $B$ are not bounded above, this means that the values of $a$ or $b$ can be arbitrarily large. Thus, the denominator $a^2 + b$ may be arbitraily large. If this is the case, $0$ is the greatest lower bound of $C$ since the fractions 
$$\frac{1}{a^2 + b}$$
become closer and closer to $0$ as the denominator gets bigger and bigger. 
