Existence, Uniqueness and Counterexamples · Hard problem

There is an $M$ such that for every $x$, $f(x) > M$.

For every $x$ there is an $M$ with $f(x) > M$.

For every $M$ and every $x$, $f(x) > M$.

For every $M$ there is an $x$ with $f(x) > M$.