Show that a limit set is connected? How do i prove this question:
Show that a limit set is connected (that is, not the union of two disjoint non- empty closed sets)
This question is in the book (Morris W. Hirsch, Stephen Smale - Differential Equations, Dynamical Systems,
and Linear Algebra)page241
 A: As stated in my comment this is not generally true. For example define the vector field
$$
f(x,y)=(y+x(1-x^2-y^2),-x+y(1-x^2-y^2))
$$
on the open set $A=\mathbb{R}^2 \setminus \{q,r\}$ where $q,r$ are two distinct points on the unit circle $C$. Then if $p \notin \{0\} \cup C \setminus \{q,r\}$ one can show that $\omega(p)=C \setminus \{q,r\}$ which is not connected.
If you suppose $\omega(p)$ is compact than we can prove this as follows:
Suppose $\omega(p)$ is not connected. Then there are $A,B$ both closed and non empty such that $\omega(p)=A\cup B$ and $A \cap B= \emptyset$. Take $a \in A$ and since $a \in \omega(p)$ there is a sequence $(s_n)$ such that $s_n \to \infty$ and $\varphi(s_n,p) \to a$ as $n \to \infty$. By the same argument take $b \in B$ and there is a sequence $(r_n)$ such that $r_n \to \infty$ and $\varphi(r_n,p) \to b$ as $n \to \infty$. Denote $d=d(A,B)$. Notice $d>0$ (why?). Using the sequences above we can define (do you see how?) a sequence $(t_n)$ such that $d(\varphi(t_n,p),A)<\frac{d}{2}$ and $d(\varphi(t_{n+1},p),A)>\frac{d}{2}$ for every $n$ odd. Hence, for all odd $n$, the function defined by $h(t)=d(\varphi(t,p),A)$ is continuous on $t_n \leq t \leq t_{n+1}$ and $g(t_n)<\frac{d}{2}$, $g(t_{n+1})>\frac{d}{2}$. By intermediate value theorem there is some $t_n<\overline{t}_n < t_{n+1}$ such that  $d(\varphi(\overline{t}_n,p),A)=g(\overline{t}_n)=\frac{d}{2}$. Notice the sequence $(\varphi(\overline{t}_n,p))$ is contained in the set S=$\{x \mid d(x,A)=\frac{d}{2}\}$. Since $S$ is compact (why?) we can find a subsequence of $(\varphi(\overline{t}_n,p))$ that converges to a point $\overline{p} \in S$ (without lost of generality assume that $\varphi(\overline{t}_n,p) \to \overline{p}$). Hence by definition $\overline{p} \in \omega(p)$. But $\overline{p} \in S$ implies that $d(\overline{p},A)=\frac{d}{2}>0$. Then $\overline{p} \notin A$. On the other hand
$$
d(A,B) \leq d(\overline{p},A)+d(\overline{p},B)
$$
$$
d(\overline{p},B)\geq d - d(\overline{p},A)=d-\frac{d}{2}=\frac{d}{2} > 0.
$$
And $\overline{p} \notin B$. This is a contradiction because $\omega(p)=A \cup B$.
