Multiple Variables in Asymptotic Notation
Suppose that we consider $f(n,m) = 1$ and $g(n,m) = n$ as functions on $[0,\infty)^2$. Assume, for the sake of contradiction, that $f(n,m) = O(g(n,m))$. According to the definition, there exist $M,C&gt;0$ such that $|f(n,m)| \leq C|g(n,m)|$ whenever $\max(n,m) \geq M$. In particular, this should hold for $(n,m)=(0,M)$, yet in this case $|f(n,m)| &gt; 1 &gt; 0 = C|g(n,m)|$.
The problem disappears if the domain is $[1,\infty)^2$. Indeed, in this case we can take $M=C=1$, since $g(n,m) = n \geq 1 = f(n,m)$ for all $n,m \in [1,\infty)^2$.
Perhaps the problem is with unpacking the definition. According to the Wikipedia definition, $f(\vec{x}) = O(g(\vec{x}))$ if there exist $M,C&gt;0$ such that the following holds:

If $x_i \geq M$ for some $i$, then $|f(\vec{x})| \leq C|g(\vec{x})|$.

Equivalently,

If $\max(\vec{x}) \geq M$, then $|f(\vec{x})| \leq C|g(\vec{x})|$.

It should be noted that this definition is not standard, and one could think of other definitions. Two particularly natural alternatives are (i) requiring all entries of $\vec{x}$ to be large, and (ii) requiring the inequality $|f(\vec{x})| \leq C|g(\vec{x})|$ to hold for all $\vec{x}$.
(The second option is related to the fact that for functions $f,g\colon \mathbb{N} \to \mathbb{N}_{&gt;0}$, the usual definition of $f=O(g)$, which is a special of the definition above, is equivalent to a definition which holds for all $n \in \mathbb{N}$.)