Question on suprema of a set $X$ which is a subset of set of real numbers

Question

Suppose, I have a non empty set $X$ that is bounded above. Also all the elements of X are non negative. How do I prove that suprema of $X$ is non negative. I am just struggling with what axioms/theorems to use or start with? I will explain where I am confused by showing my attempt:

I tried proving by contradiction. First, since $X$ is bounded above, completeness axiom says that $\sup X$ does exist and assumed that $\sup X < 0$. But by approximation theorem, we have, for a bounded above set $X$, for any positive $e > 0$, there exists a $x$ belonging to set $X$ such that

$$\sup X - e < x \leq \sup X$$

Now if $\sup X < 0$, by the second inequality, any $x$ belonging to set $X$ is also negative which is in contradiction to the given fact that $X$ has non negative elements.

Is this a good way to go about proving it?

Any suggestion would be very helpful.

Answer 1

It follows easier without contradiction. $\sup X$ is an upper bound for $X$. Pick $x \in X$. Then $x \geq 0$, so $\sup X \geq x \geq 0$.

Answer 2

You could prove it by contradiction... but I prefer to do it directly...

Let $x\in X$. By assumption $x\gt 0$... So $supX\ge x\gt 0$...

Your proof has an error in it... You don't need $\epsilon $. By definition $x\le supX\lt 0$. Contradiction.