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Consider a first order autonomous ODE:

$$\frac{dy}{dt}=f(y)$$

The solutions to this ODE cannot cross each other due to the Picard-Lindelöf theorem (lets say this ODE satisfies it). Also note that, because the sign of $f(y)$ can only change at the zeros and the function can't cross the zeros since they are (equilibrium) solutions, $f(y)$ is monotone.

These two reasons put together mean that any solution to a 1st order autonomous ODE are bounded by the equilibrium solutions that surround them:

enter image description here

p and q are equilibrium solutions, i.e. f(p)=f(q)=0.

This is what my introductory textbook on differential equations and all the online resources I could find say about the subject. What I don't get is how they go further and say that the limit of these bound solutions approach the equilibria as $t\to\infty$ or $-\infty$ depending on if $f(y)$ is decreasing or increasing or, if there is no bounding equllibrium solution, $f(y)$ approaches $\pm\infty$.

I mean I believe them, and it certainly seems intuitive, yet I don't see how monotonicity and not being able to cross over guarantee this alone. Couldn't you have a solution with a horizontal asymptote slightly lower/higher than the next highest/lowest equilibria? Such a solution would still be monotone, and still be bounded by the equilibrium solutions, yet doesn't have to get close to them:

enter image description here

So why must these solutions approach equilibrium points/explode to infinity if there are no available equilibrium points?

Edit: looking at the slope field of my proposed solution, I see that the answer to my question probably has to do with the time invarience of these solutions as the slope field can't change over time to make my proposed solution possible without the asymptote also being an equilibrium solution. But how do I make this precise instead of a graphical intuition?

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The sequence $(y(n))_{n\in\Bbb N}$ is bounded and increasing. By basic calculus, it means that it converges. As $y$ is monotonous, all the other function values between the integers converge with this sequence. But $$y(n+1)-y(n)=y'(n+\theta_n)=f(y(n+\theta_n))$$ converges to $0$ which is only possible if $y(n+\theta_n)$ converges to a stationary point $y_*$, $f(y_*)=0$, which means $y(t)\to y_*$ for $t\to\infty$.

Similar arguments apply for the other direction.

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