Given the system of ODEs $$x' = x(1-y)$$ $$y'=y(x-1),$$ let the set $Q=\{(x,y):x\ge 0, y\ge 0\}$. Explain why $Q$ is invariant for this system of ODEs.
My explanation: If $x > 1$ and $y<1$ then $x'$ and $y'$ are both greater than $0$, which means that the solution stays in $Q$. If $y>1$ or $x\le1$ then $x'$ or $y'$ is/are negative, so the solution is decreasing towards at least one of the axes. But since both axes are invariant under the flow of the given system, the flow stays in $S$.
Please let me know if this is correct.
Since the axes are invariant, any connected component of the complement of their union is also invariant. That is, each open quadrant is invariant.