Can the $\min$ function take countably infinitely many arguments?

Question

Can the $\min$ function take countably infinitely many arguments?

That is, does it make sense to have $\min(a_1, a_2, ...)$?

Answer 1

If it could, what would be $\min(0,1,-1,2,-2,\ldots)$? In fact, what would be $\min(1,\frac12,\frac13,\frac14,\ldots)$?

For infinite sets we no longer guaranteed to have a minimum. But we do have an infimum, at least as far as bounded subsets of the real numbers are involved. So we can talk about $\inf(1,\frac12,\frac13,\ldots)=0$.

Note, by the way, that the $\inf$ function actually takes a set, not just a countable family of arguments. We can talk about $\inf(\{x\mid x^3>2\})=\sqrt[3]2$, for example. But as luck would have it, we can show that every set can be reduced to a countable sequence which converges to its infimum.

Answer 2

An infinite min only makes sense if you are dealing with subsets of a well ordered set. Well ordered sets have the property that any subset has a unique minimum element.

For example $\mathbb{N}$ is well ordered as is any ordinal. Well ordered sets are useful because they allow you to do induction on them.