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.

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.
