Definition
Let $\mathcal{L}$ be a non-empty collection of real-valued functions on a set $X$. Then $\mathcal{L}$ is a real vector space iff for all $f,g\in\mathcal{L}$ and $c\in\mathbb{R},cf+g\in\mathcal{L}$. Let $f\lor g:=\mathrm{max}(f,g),f\land g:=\mathrm{min}(f,g)$. A vector space $\mathcal{L}$ of functions is called a vector lattice iff for all $f$ and $g$ in $\mathcal{L}$, $f\lor g\in\mathcal{L}$. Then also $f\land g\equiv-(-f\lor-g)\in\mathcal{L}$.
Problem
Let $f$ be a function on a set $X$ and $\mathcal{L}:=\{cf:c\in\mathbb{R}\}$. Show that $\mathcal{L}$ is a vector lattice if $f\geq0$ or if $f\leq0$ but not otherwise.
My effort
$\mathcal{L}$ is a real vector space since for all $c_1f,c_2f\in\mathcal{L}$ and $c\in\mathbb{R},cc_1f+c_2f=(cc_1+c_2)f\in\mathcal{L}$.
If $f\geq0$, then $c_1f\lor c_2f=\mathrm{max}(c_1,c_2)f\in\mathcal{L}$. Thus $\mathcal{L}$ is a vector lattice.
If $f\leq0$, then $c_1f\lor c_2f=\mathrm{min}(c_1,c_2)f\in\mathcal{L}$. Thus $\mathcal{L}$ is a vector lattice.
My question
I think the "but not otherwise" is wrong. Let $\mathcal{L}$ be a vector lattice. Then $f\lor-f\in\mathcal{L}$, i.e., there exists $c\in\mathbb{R}$ such that $f\lor-f=cf$. Assume neither $f\geq0$ nor $f\leq0$ is true. Then exist $x_1,x_2\in X$ such that $f(x_1)>0$ and $f(x_2)<0$. So we have $(f\lor-f)(x_1)=f(x_1)=cf(x_1)$ and $(f\lor-f)(x_2)=-f(x_2)=cf(x_2)$. But this requires $c=1$ and $c=-1$ at the same time, which is impossible. Therefore, it must be either $f\geq0$ or $f\leq0$.
Where am I wrong?
If $f$ is neither non-positive or non-negative, then $$ f\vee (-f)\ge 0, $$ which means that $f\vee (-f)\notin \mathcal{L}$ ($\because$ there is no $c>0$ s.t. $cf=f\vee (-f)$ unless $f\equiv 0$).