The pointwise order on functions is partial

Statement

For $f, g : \mathbb{R} \to \mathbb{R}$ set $f \preccurlyeq g$ if and only if $f(x) \le g(x)$ for every $x \in \mathbb{R}$. Prove that $\preccurlyeq$ is a (partial) order relation. Is it a total order?

Solution

Reflexive: $f(x) \le f(x)$ for every $x$, so $f \preccurlyeq f$. Antisymmetric: if $f \preccurlyeq g$ and $g \preccurlyeq f$, then for every $x$ we have $f(x) \le g(x)$ and $g(x) \le f(x)$, so $f(x) = g(x)$ for all $x$, i.e. $f = g$. Transitive: if $f \preccurlyeq g$ and $g \preccurlyeq h$, then for every $x$ we have $f(x) \le g(x) \le h(x)$, so $f \preccurlyeq h$.

All three properties hold, so $\preccurlyeq$ is a partial order.

It is not total. Take $f(x) = x$ and $g(x) = -x$. At $x = 1$: $f(1) = 1 > -1 = g(1)$, so $f \not\preccurlyeq g$. At $x = -1$: $f(-1) = -1 < 1 = g(-1)$, so $g \not\preccurlyeq f$. The two functions are incomparable. $\blacksquare$