# If omega limit set contains only one point, $x^*$, then $\lim_{t\to\infty}\phi(t;x_0)=x^*$

I'm trying to proof the following, and I'm looking for a verification of my proof. If it is incorrect, I'm looking for some help towards a right proof.

If the omega limit set is $$\omega(x_0)=\{x^*\}$$, then $$\lim_{t\to\infty}\phi(t;x_0)=x^*$$

Suppose $$\lim_{t\to\infty}\phi(t;x_0)$$ exists and equals $$x'$$. Then $$\omega(x_0)=\{x'\}$$, so it suffices to proof that the limit exists. Now assume for the sake of contradiction that $$\lim_{t\to\infty}\phi(t;x_0)$$ does not exist. Because $$x^*\in\omega(x_0)$$, there exists a sequence $$(t_n)_{n\in\mathbb{N}}$$ such that $$\lim_{n\to\infty}\phi(t_n;x_0)=x^*$$. Because $$\lim_{t\to\infty}\phi(t;x_0)$$ does not exist, there exists a $$\delta>0$$ such that there is no $$T$$ such that $$\phi(t;x_0)\in B_{\delta}(x^*)$$ for all $$t\geq T$$ (otherwise the limit would exist). Fix this $$\delta$$. From continuity of the flow and existence of a sequence $$(t_n)_{n\in\mathbb{N}}$$ satisfying $$\lim_{n\to\infty}t_n=\infty$$ and $$\lim_{n\to\infty}\phi(t_n,x_0)=x^*$$, it follows that for an arbitrary $$\epsilon>\delta$$, there exists a sequence $$(s_n)_{n\in\mathbb{N}}$$ satisfying $$\lim_{n\to\infty}s_n=\infty$$ such that for all $$n\in\mathbb{N}$$, $$\phi(s_n;x_0)\in\overline{B_{\epsilon}(x^*)}\setminus B_{\delta}(x^*)$$, which is in particular a closed set. Now it follows from Bolzano-Weierstrass that $$\left(\phi(s_n;x_0)\right)_{n\in\mathbb{N}}$$ has a convergent subsequence $$\left(\phi(s_{n_k};x_0)\right)_{k\in\mathbb{N}}$$, belonging to $$\overline{B_{\epsilon}(x^*)}\setminus B_{\delta}(x^*)$$, because this set is closed, and thus the limit value is unequal to $$x^*$$; thus $$\omega(x_0)\neq\{x^*\}$$, which gives the desired contradiction.

Edit, in response to the comment: "Would you have any shorter proof in mind or do you think this is one of the simplest possible?".

This is the simplest proof that I know of. The presentation can be made very easy to read too.

Lemma. If $$K$$ is sequentially compact and $$\phi(t_n;x_0)\in K$$ for times $$t_n\to\infty$$, then $$\omega(x_0)\cap K\neq\emptyset$$.

Proof. Immediate just from the definitions.

If, by contradiction, $$x^*$$ is not the limit of $$\phi(t;x_0)$$, there is $$\delta>0$$ and $$t_n\to\infty$$ such that $$\phi(t_n;x_0)\in B_\delta(x^*)^c$$.

Since $$x^*\in\omega(x_0)$$, there are times $$t'_n\to\infty$$ such that $$\phi(t'_n;x_0)\in B_{\delta/2}(x^*)$$.

By continuity of $$\phi(\,\cdot\,;x_0)$$, there are times $$t''_n\to\infty$$ such that $$\phi(t''_n;x_0)\in \overline{B_\delta(x^*)}\setminus B_{\delta/2}(x^*)=K$$. Then the Lemma implies that $$\omega(x_0)$$ cannot be just $$\{x^*\}$$.

• Thanks for reading. Would you have any shorter proof in mind or do you think this is one of the simplest possible? – Václav Mordvinov Oct 29 '18 at 18:06
• Thank you for your edit. Indeed the idea is the same, but it is way shorther and better readable – Václav Mordvinov Oct 31 '18 at 7:06
• @LutzL you are totally right! thanks for spotting the typo – Federico Nov 5 '18 at 13:02