Prove or give a counterexample

Question

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.

Answer 1

It's wrong!

Let be it true.

Thus, $t=s+1$ gives a counterexample.

Answer 2

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.

Answer 3

A non-negative number that is greater than or equal to all non-negative numbers I think does not exist.

Answer 4

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.