I am trying to understand how one can prove the following assertion using a continuity argument:


Let $0<\epsilon<\epsilon\_0$. Let $I=[t\_0,R]$ be a compact interval. Suppose that $S:I\to [0,\infty)$ is a continuous non-decreasing function such that $S(t\_0)=0$ and
$$S(T)\lesssim \epsilon\_0(S(T)+\epsilon)^4+\epsilon\_0^4(S(T)+\epsilon)+\epsilon\_0(S(T)+\epsilon)^5$$
for all $T\in I$. (Here, $a\lesssim b$ means $a\le Cb$ for some constant $C>0$) Then if $\epsilon\_0$ can be chosen to be sufficiently small, we in fact have
$S(T)\le \epsilon$ for all $T\in I$.


From my understanding, the idea of continuity argument here is to consider the set $\Omega=\{T\in I: S(T)\le \epsilon\}$. Then $\Omega$ is closed and non-empty since $t\_0\in\Omega$. If we can prove that $\Omega$ is also open (which I am stuck on) then the connectedness of $I$ would imply that $\Omega=I$, which is what we need.


As for the context of the problem, I am stuck at reading this paper by Tao and Visan (Proof of Theorem 1.3, page 17, the assertion is labelled as (3.6))
<http://arxiv.org/abs/math/0507005>


Apologies if this is actually trivial. I am completely new to this method of continuity argument.



A continuity/bootstrap argument
I tend to think about such continuity argument in this way:


Let $T^\*=\sup\{T\in I: S(t)\le\epsilon\}$ for all $t\le T$. Then it suffices to show $T^\*=R$.


Suppose not. Then by continuity we have $S(T^\*)=\epsilon$. Plugging this into the given inequality we get $\epsilon\ll \epsilon\_0(\epsilon^4+\epsilon^5)+\epsilon\_0^4\epsilon$. Then it is clear that if $0<\epsilon<\epsilon\_0$ is small enough, then this inequality cannot hold.


