But if there is some $\epsilon_0>0$ such that $$\sup(A+B)+\epsilon_0=\sup A+\sup B,$$ then there is some $\epsilon>0$ with $$\sup(A+B)+\epsilon<\sup A+\sup B$$ (for example, $\epsilon=\frac{\epsilon_0}2$ works), so we can't show that $$\sup(A+B)+\epsilon\ge\sup A+\sup B$$ for all $\epsilon>0$, since it's not true. Perhaps I don't understand what you're asking, though. When you say "That's not true then," what is the word that referring to?

Take a look at my updated answer, and see if it makes more intuitive sense to you.

Nothing is wrong at all. In the case $x=4,y=5$, we can do exactly as I suggested in my proof above: let $z$ be the midpoint of $x$ and $y$--namely, $4.5$--which gives us a real number bigger than $x$ that is not at least as big as $y$. The word "every" is a key word, here. No matter what real numbers $x$ and $y$ are, there will always be some real number bigger than $x$ that is at least as big as $y$--for example, $\max(x,y)+1$--so a single example (indeed, even infinitely many examples) will not prove that $x\ge y$. We need to show it for all real numbers bigger than $x$.

I recommend trying to understand why and how it works, first. That should help you to have a better idea when it may be useful. Certainly in situations like this, when we're dealing with suprema and infima, this sort of trick may come in handy. Also, it's a fair bet that when it's given as a hint, it may be worth using.

No. It does not imply $5>5$. Why do you think you can conclude that?

Ah! I see where you're getting confused. The word "if" is another key word. "If every real number bigger than $x$ is at least as big as $y$, then $x$ is at least as big as $y$." The first condition fails to be met in the case that $x=4,y=5$ (since $4.5$ is a counterexample), so we cannot conclude from the above-quoted statement that $x\ge y$ in that case.

Let me give you a further example to illustrate the point: "If every positive real number is at least as big as $y$, then $y$ is nonpositive." Letting $y=1$, the positive real number $\frac12$ is not at least as big as $y$, so that statement does not imply that $y$ is nonpositive. It's true that (for example) the positive real number $2$ is at least as big as $y$, but since there are positive reals that aren't at least as big as $y$, then the first condition fails, and so the conclusion does not follow.

Since no real number is smaller than itself, then there is certainly no such thing as "a (real) number smaller than every real number". That's not what I said, though. Let me rephrase slightly for clarity: "If the real numbers bigger than $x$ are all at least as big as $y$, then $x$ is at least as big as $y$." For example, every number bigger than $\pi$ is at least as big as $3$. Hence, we can conclude that $\pi$ is at least as big as $3$.

Another alternative, put into the language of sets and subsets: "If $(x,\infty)\subseteq[y,\infty),$ then $[x,\infty)\subseteq[y,\infty)$."

may I prove $\star$ as follows? $y \leq x \implies y + \epsilon \leq x + \epsilon; y \leq y + \epsilon \leq x + \epsilon$. On the other hand, we have $\lim_{\epsilon \to 0} y \leq \lim_{\epsilon \to 0} (x + \epsilon) = x$

@sidht: Your first line looks like you're assuming what you want to prove. Can you clarify what you're assuming?

I want to prove $y \leq x \iff y \leq x + \epsilon$. Or I guess we could turn the inequality on the RHS to a strict inequality.

Ah! I see. Indeed, then $y\le x\implies y\le x+\epsilon$ for all $\epsilon>0$, exactly as you described. You'll need to be a bit cautious about using limits to go the other way, since our definition of limits already uses an $\epsilon$, and uses $x$ as a variable! It might be better to proceed as follows: (1) Note that $0$ is a lower bound of the set of positive reals (by definition of positive). (2) Note/prove that half of a positive number is a smaller positive number, thus proving that $0$ is the infimum of the set of positive reals. (3) Suppose $y\le x+\epsilon$ for all $\epsilon>0.$ ...

... But that means $y-x\le\epsilon$ for all $\epsilon>0,$ meaning $y-x$ is a lower bound of the positive numbers, so since $0$ is the greatest lower bound of the positive numbers, we have $y-x\le 0,$ and so $y\le x$.

Sorry I am still confused as to what is wrong with the limits. There is something wrong with it, but I don't think it is the reason you gave. When i took the limit, isn't $x$ a constant?

It is, certainly, but in our definition, it is not, so there's potential for confusion. (Note also that we want the limit as $\epsilon\to 0^+$). To avoid that, we can rewrite our definition, and show that for all $c>0,$ there is some $d>0$ such that $$0<\epsilon-0<d\implies|(x+\epsilon)-x|<c,$$ or equivalently, $$0<\epsilon<d\implies|\epsilon|<c.$$ But this is simple. Take any $c>0,$ and put $d=c,$ so that for $0<\epsilon<d,$ we have $|\epsilon|=\epsilon<d=c.$ The approach works just fine. I merely wanted to advise caution.