Im stuck on how to prove this:
$$\inf(x) + \sup(y) \leq \sup(x+y)$$
I know that
$$\inf(x) \leq \sup(x)\\ x \leq y\implies \sup(x) \leq \inf(y)\\ \sup(x) + \sup(y) \geq \sup(x+y)\\ \inf(x) + \inf(y) \leq \inf(x+y)\\ x \leq y\implies \sup(x)\leq \sup(y)$$
How can I utilize these to prove this?
I've come to the conclusion that $$\inf(x)+\inf(y) \leq \inf(x+y) \leq \sup(x+y) \leq \sup(x) + \sup(y)$$
I've also tried utilizing the $\sup(x) = -\inf(-x)$ proof, but that doesn't seem to work either.
We can prove so by definition. For any $\epsilon>0$, there is $u\in y$ so that $$\sup(y) - \epsilon < u$$
For this $u$, pick any $v\in x$, $$\inf(x) +\sup(y) - \epsilon < v + u \le \sup(x+y).\tag{1}$$ Since (1) holds for any $\epsilon > 0$, we have $$\inf(x) + \sup(y) \le \sup(x+y).$$