Prove or give a counterexample 
There exists a non-negative number s such that for all non-negative numbers t, the inequality s $\geq$ t holds

Don't really know if this statement is true or false. I can see it being either, but can't really approach a proof or a counterexample for it.
Thanks.
 A: It's wrong!
Let be it true.
Thus, $t=s+1$ gives a counterexample. 
A: I don't think so. $s+1>s$ for all non negative $s$.
The above example or whatever you call it works if you are talking about an infinite set. For finite set, it is possible to have that $s$. For instance in the set $\{1,2,3,4,5,6,7,8,9\}$ you have your desired element.
A: A non-negative number that is greater than or equal to all non-negative numbers I think does not exist.
A: The quantifiers are in the order "first $s$, then $t$".  Since $t$ may depend on $s$, we should think of $t$ as a function of $s$.  Can we write constraints for $t$ in terms of $s$ that make the inequality true?
We find that $0 \leq t$ because $t$ is nonnegative and $s \geq t$ is our inequality, so we must have $t \in [0,s]$ for the inequality to hold.
Can we pick an $s$ so large that all possible choices of $t$ are in $[0,s]$?  No.  If we let $t = s + 1$ (or let $t$ be any function of $s$ that is always bigger than $s$), then $t \not\in [0,s]$, so not $s \geq t$.
Note: We needed to write $t$ as a function of $s$ in the above, so that we simultaneously showed each choice for $s$ does not work.
