Prove that sup $f(x)$ $\le$ inf $g(y)$. (proof verification)

Question

I saw this proof, but I don't understand one point.

$f(x) \le g(y)$ for $x,y \in D$

$\rightarrow$ sup$f(x)$ $\le$ $g(y)$

$\rightarrow$sup$f(x)$ $\le$ inf$g(y)$

I think from the second line to third line of proof, it assumes that inf$g(y)$ $\in$ $g(D)$.

But, how can we prove this?

Thanks in advance.

Answer 1

Since $f(x)\le g(y)$ for $x$,$y\in D$

$g(y)$ is an upper bound for $f(x)$, so the least upper bound is less than or equal $g(y)$

Thus $\sup f(x) \le g(y)$. That implies $\sup f(x)$ is a lower bound for $g(y)$ so the greatest lower bound of $g(y)$ is greater than or equal to sup $f(x)$.

That is:

$\sup f(x) \le \inf g(y)$

Answer 2

The property that they use is let $a \in \mathbb{R}$. Then $\forall y, a \le g(y) \implies a \le \inf_y g(y)$.

$a$ is a lower bound for the set $\{ g(y): y \in D\}$, but remember that the infimum is the greatest lower bound, hence we have $a \le \inf_y g(y)$.

Here $a$ is $\sup f(x)$.

Answer 3

Take any $\epsilon > 0$, then there exists $x$ such that

$$\sup f - f(x) < \epsilon $$

similarly there exists a $y$ such that

$$ g(y) - \inf g < \epsilon$$

Then

$$\sup f < \epsilon + f(x) \leq \epsilon + g(y) <\epsilon +\left( \inf g + \epsilon\right) = \inf g + 2\epsilon$$

As $\sup f < \inf g + 2\epsilon$ for all $\epsilon > 0$, letting $\epsilon \to 0^+$ gives $$\sup f \leq \inf g$$

Answer 4

You do not need to assume that $\inf g\in g(D)$. Assume by way of contradiction that $\sup f> \inf g$. Then there exists some $\delta>0$ such that $\sup f>\inf g+\delta$. By definition of the infimum there exists some $y\in D$ such that $\inf g\leq g(y)\leq \inf g +\delta$. Thus $g(y)<\sup f$, contradicting part $2$ of your proof.