Consider a system of ODEs of the form $$\dot x_1 = f_1(x_1,x_2)-g_1(x_1,x_2)x_1 \\ \dot x_2 = f_2(x_1,x_2)-g_2(x_1,x_2)x_2,$$where $f,g$ are bounded, Lipschitz-continuous functions. (Then by the existence and uniqueness theorem, solutions exist - even more, by boundedness of $f,g$ they exist on $\mathbb{R}$.)
My question is: Do these assumptions of $f$ and $g$ imply that there exists an closed, invariant cube, $[0,K]\times [0,K]$, for some $K>0$ ? If yes, does this cube contain all attractors of the system ?
Assuming suitable positivity constraints for $f_k$ and $g_k$, the vector field points towards the inside of the first quadrant on the positive half axes.
If the $g_k$ are bounded away from zero, then for a large enough box attached to the coordinate axes the vector field will point inwards also on the other two sides of the box.
Then apply the Poincare-Bendixon theorem.
Details
If $0<f_k<M$ and $n<g_k$ then $f_k(x)-g_(x)x_k<M-nx_k\le0$ for $x_k\ge K=M/n$.
On the other sides one has $f_1(0,x_2)-g_1(0,x_2)·0>0$ and $f_2(x_1,0)-g_2(x_1,0)·0>0$.
Thus $F=(f_1-g_1x_1,f_2-g_2x_2)$ points upwards on the $x_1$-axis, right on the $x_2$-axis, down on $x_2=K$ and left on $x_1=K$.
One needs the positive lower bound for $g$ because otherwise one can construct examples where $g_1(x)=1/(1+x_1)$. Then if $f_1(x)>1$, the difference $f_1-x_1/(1+x_1)$ is always positive, allowing for infinite growth of the solutions.