supremum: some relations about concluding step I  don't understand
$$ a > \sup ( A ) - \frac \epsilon 2 $$
$$ b > \sup ( B ) - \frac \epsilon 2 $$
$$ a + b > \sup ( A ) + \sup ( B ) - \epsilon $$
how do we conclude that $ \sup ( A ) + \sup ( B ) \le \sup ( A + B ) $?
I also want to ask how to study this course; can you recommend anything because I am struggling this course?
 A: We have to use the following theorem, which is used in Analysis very often, and usually without mention:
Theorem: Let $x, y$ be real numbers, and suppose that for every $\epsilon > 0$, $$x \ge y - \epsilon.$$
Then, $x \ge y$.
Proof: (Informal) How do we prove this? Contradiction is always good to try. Assume $x < y$, and imagine them on the number line. It looks like this:
$----x --- (gap) --- y -----$
$x$ is to the left of $y$. So if we can find an $\epsilon^*$ such that $y - \epsilon^*$ is in the gap, then we can get a contradiction, because $y - \epsilon^* > x$. How can we find such an $\epsilon^*$? We can find a number $z^*$ in the gap, and then we can take $\epsilon^* = y - z^*$, and it should work. How can we find a number $z^*$, in the gap between $x$ and $y$? Well the average of $x$ and $y$, $(x + y)/2$ should be in that gap, so we can use $\epsilon^* = y - \frac{x+y}{2} = \frac{y-x}{2}.$
As an exercise, you should write this proof formally.
Now, for the main question:
We showed $a + b > \sup ( A ) + \sup ( B ) - \epsilon$. For every $\epsilon > 0$. Now $\sup(A+B)$ is supposed to be $\ge$ than any element in $A+B$, so in particular $\sup(A+B) \ge a + b$ for this $a$ and $b$. Thus
$$\sup(A+B) \ge \sup(A) + \sup(B) - \epsilon.$$
Now apply the theorem.
