[proofplan] We prove each property of $\omega(x)$ separately. Non-emptiness follows from the Bolzano–Weierstrass theorem applied to the bounded forward orbit. Closedness is immediate from the definition of $\omega(x)$ as an intersection of closed sets. Boundedness follows from the boundedness of $\mathcal{O}^+(x)$. Invariance under the flow uses continuous dependence of solutions on initial data. Connectedness is proved by contradiction: if $\omega(x)$ were disconnected, separation by open sets would force the forward orbit to eventually avoid one component, contradicting the definition. [/proofplan] [step:Establish non-emptiness via the Bolzano–Weierstrass theorem] Since $\mathcal{O}^+(x) = \{\varphi(t, x) : t \geq 0\}$ is bounded in $\mathbb{R}^n$, any sequence $(\varphi(t_k, x))_{k \geq 1}$ with $t_k \to +\infty$ is a bounded sequence in $\mathbb{R}^n$. By the [Bolzano–Weierstrass Theorem](/theorems/???), it has a convergent subsequence $\varphi(t_{k_j}, x) \to y$ for some $y \in \mathbb{R}^n$. By definition, $y \in \omega(x)$. Hence $\omega(x) \neq \varnothing$. [/step] [step:Show $\omega(x)$ is closed and bounded] Recall that $\omega(x) = \bigcap_{T \geq 0} \overline{\{\varphi(t, x) : t \geq T\}}$. Each set $\overline{\{\varphi(t, x) : t \geq T\}}$ is closed (it is a closure), so $\omega(x)$ is an intersection of closed sets, hence closed. For boundedness: since $\mathcal{O}^+(x)$ is bounded, there exists $R > 0$ with $|\varphi(t, x)| \leq R$ for all $t \geq 0$. Then $\overline{\{\varphi(t, x) : t \geq T\}} \subset \overline{B}(0, R)$ for every $T \geq 0$, so $\omega(x) \subset \overline{B}(0, R)$. [guided] The set $\omega(x)$ consists of all accumulation points of the forward orbit as $t \to +\infty$. There is an equivalent characterisation as an intersection of closures: \begin{align*} \omega(x) = \bigcap_{T \geq 0} \overline{\{\varphi(t, x) : t \geq T\}}. \end{align*} To see this equivalence: $y \in \omega(x)$ means there exist $t_k \to +\infty$ with $\varphi(t_k, x) \to y$. For any $T \geq 0$, all but finitely many $t_k$ satisfy $t_k \geq T$, so $y \in \overline{\{\varphi(t, x) : t \geq T\}}$. Conversely, if $y$ lies in every such closure, then for each integer $k \geq 1$, there exists $t_k \geq k$ with $|\varphi(t_k, x) - y| < 1/k$, giving $t_k \to +\infty$ and $\varphi(t_k, x) \to y$. Each $\overline{\{\varphi(t, x) : t \geq T\}}$ is closed by definition, so $\omega(x)$ is an intersection of closed sets and therefore closed. Boundedness follows because $\mathcal{O}^+(x)$ lies in some ball $\overline{B}(0, R)$, and the closure of a bounded set is bounded. [/guided] [/step] [step:Prove invariance under the flow using continuous dependence on initial data] We must show that for every $y \in \omega(x)$ and every $s \in \mathbb{R}$, $\varphi(s, y) \in \omega(x)$. Let $y \in \omega(x)$. Then there exist $t_k \to +\infty$ with $\varphi(t_k, x) \to y$. By continuous dependence of solutions on initial data (the [Picard–Lindelöf Theorem](/theorems/2774) guarantees this), the map $z \mapsto \varphi(s, z)$ is continuous for each fixed $s$. Therefore \begin{align*} \varphi(s, \varphi(t_k, x)) \to \varphi(s, y) \quad \text{as } k \to \infty. \end{align*} By the flow property (semigroup law), $\varphi(s, \varphi(t_k, x)) = \varphi(t_k + s, x)$. Since $t_k \to +\infty$, we have $t_k + s \to +\infty$ as well (for any fixed $s \in \mathbb{R}$). Hence $\varphi(s, y)$ is a limit of $\varphi(t_k + s, x)$ along a sequence of times tending to $+\infty$, which means $\varphi(s, y) \in \omega(x)$. Since $s \in \mathbb{R}$ was arbitrary, $\omega(x)$ is invariant under the flow: $\varphi(s, \omega(x)) = \omega(x)$ for all $s$. [guided] We must show: for every $y \in \omega(x)$ and every $s \in \mathbb{R}$, the point $\varphi(s, y)$ also belongs to $\omega(x)$. The argument uses two ingredients: the semigroup law for the flow and continuous dependence of solutions on initial data. Why does invariance require continuous dependence on initial data, rather than just continuity of the flow in $t$? Because the limit $y = \lim_{k \to \infty} \varphi(t_k, x)$ is an initial condition for a new IVP, not a time limit along a fixed solution. We need to know that $\varphi(s, \cdot)$ is continuous as a function of the initial point: starting near $y$ and flowing for time $s$ should end near $\varphi(s, y)$. Fix $s \in \mathbb{R}$. Since $y \in \omega(x)$, there exist $t_k \to +\infty$ with $\varphi(t_k, x) \to y$. The [Picard-Lindelof Theorem](/theorems/2774) guarantees that $\varphi(s, z)$ depends continuously on the initial condition $z$ for each fixed $s$. Applying this continuous dependence: \begin{align*} \varphi(s, \varphi(t_k, x)) \to \varphi(s, y) \quad \text{as } k \to \infty. \end{align*} By the semigroup (flow) property, $\varphi(s, \varphi(t_k, x)) = \varphi(t_k + s, x)$. Since $t_k \to +\infty$, for any fixed $s \in \mathbb{R}$ we have $t_k + s \to +\infty$ as well (for all sufficiently large $k$, $t_k + s > 0$, so $\varphi(t_k + s, x)$ is well-defined as a point on the forward orbit). Therefore $\varphi(s, y)$ is a limit of $\varphi(t_k + s, x)$ along a sequence of times tending to $+\infty$, which places $\varphi(s, y) \in \omega(x)$ by definition. The argument works for negative $s$ as well: even though the forward orbit $\varphi(t, x)$ is defined only for $t \geq 0$, the point $y \in \omega(x)$ lies in the domain of $f$ and the flow through $y$ exists for both positive and negative times (at least locally, by existence and uniqueness). For $t_k$ large enough, $t_k + s > 0$ even when $s < 0$, so $\varphi(t_k + s, x)$ is well-defined. Since $s \in \mathbb{R}$ was arbitrary, we conclude that $\varphi(s, \omega(x)) \subset \omega(x)$ for all $s$. The reverse inclusion $\omega(x) \subset \varphi(s, \omega(x))$ follows by applying the same argument with $-s$: if $y \in \omega(x)$, then $\varphi(-s, y) \in \omega(x)$, and $y = \varphi(s, \varphi(-s, y)) \in \varphi(s, \omega(x))$. Hence $\varphi(s, \omega(x)) = \omega(x)$ for all $s \in \mathbb{R}$. [/guided] [/step] [step:Prove connectedness by contradiction] Suppose $\omega(x)$ is disconnected. Then there exist open sets $U, V \subset \mathbb{R}^n$ with $\omega(x) \subset U \cup V$, $\omega(x) \cap U \neq \varnothing$, $\omega(x) \cap V \neq \varnothing$, and $\omega(x) \cap U \cap V = \varnothing$. Since $\omega(x)$ is compact (closed and bounded in $\mathbb{R}^n$), we may assume $U$ and $V$ are disjoint: replace $U$ by a neighbourhood of $\omega(x) \cap U$ that is disjoint from $\overline{V} \cap \omega(x)$, and similarly for $V$. The set $W := \mathbb{R}^n \setminus (U \cup V)$ is closed and disjoint from $\omega(x)$. Since $\omega(x) \subset U \cup V$ and $W$ is closed with $\omega(x) \cap W = \varnothing$, the definition of $\omega(x)$ as $\bigcap_{T \geq 0} \overline{\{\varphi(t, x) : t \geq T\}}$ implies that there exists $T_0 \geq 0$ such that $\varphi(t, x) \in U \cup V$ for all $t \geq T_0$. (If not, there would exist $t_k \to +\infty$ with $\varphi(t_k, x) \in W$; by compactness a subsequence converges to a point in $W \cap \omega(x)$, contradicting $\omega(x) \cap W = \varnothing$.) For $t \geq T_0$, define $\chi: [T_0, \infty) \to \{0, 1\}$ by $\chi(t) = 0$ if $\varphi(t, x) \in U$ and $\chi(t) = 1$ if $\varphi(t, x) \in V$. Since $\varphi(\cdot, x)$ is continuous and $U, V$ are disjoint open sets, $\chi$ is continuous, hence constant on $[T_0, \infty)$ (as $[T_0, \infty)$ is connected). Suppose $\chi \equiv 0$, so $\varphi(t, x) \in U$ for all $t \geq T_0$. Then $\omega(x) \subset \overline{U}$, which implies $\omega(x) \cap V = \varnothing$ (since $U \cap V = \varnothing$ and $\omega(x) \subset \overline{U}$ forces accumulation points to lie in $\overline{U}$, not in $V$). This contradicts $\omega(x) \cap V \neq \varnothing$. Similarly, $\chi \equiv 1$ contradicts $\omega(x) \cap U \neq \varnothing$. Hence $\omega(x)$ is connected. [guided] Suppose for contradiction that $\omega(x)$ is disconnected. Then there exist open sets $U, V \subset \mathbb{R}^n$ with $\omega(x) \subset U \cup V$, $\omega(x) \cap U \neq \varnothing$, $\omega(x) \cap V \neq \varnothing$, and $U \cap V = \varnothing$ (we may arrange disjointness because $\omega(x)$ is compact, so the two components can be separated by disjoint open neighbourhoods). Set $W := \mathbb{R}^n \setminus (U \cup V)$, which is closed and satisfies $\omega(x) \cap W = \varnothing$. **Part 1: The trajectory eventually enters $U \cup V$.** We claim there exists $T_0 \geq 0$ such that $\varphi(t, x) \in U \cup V$ for all $t \geq T_0$. If not, there would exist a sequence $t_k \to +\infty$ with $\varphi(t_k, x) \in W$. The forward orbit lies in a compact set (since $\mathcal{O}^+(x)$ is bounded), so by the Bolzano-Weierstrass theorem a subsequence converges to some $z \in W$. Since $t_k \to +\infty$, we have $z \in \omega(x)$ by definition. But $z \in W$ and $\omega(x) \cap W = \varnothing$ -- contradiction. Hence such $T_0$ exists. **Part 2: The trajectory stays in one component.** For $t \geq T_0$, define $\chi: [T_0, \infty) \to \{0, 1\}$ by $\chi(t) = 0$ if $\varphi(t, x) \in U$ and $\chi(t) = 1$ if $\varphi(t, x) \in V$. This is well-defined because $\varphi(t, x) \in U \cup V$ and $U \cap V = \varnothing$. Moreover, $\chi$ is continuous: the preimage $\chi^{-1}(\{0\}) = \{t \geq T_0 : \varphi(t, x) \in U\}$ is open in $[T_0, \infty)$ because $U$ is open and $\varphi(\cdot, x)$ is continuous, and similarly for $\chi^{-1}(\{1\})$. Since $[T_0, \infty)$ is connected and $\chi$ takes values in the discrete space $\{0, 1\}$, $\chi$ must be constant. **Part 3: Contradiction.** Suppose $\chi \equiv 0$, so $\varphi(t, x) \in U$ for all $t \geq T_0$. Then every accumulation point of $\{\varphi(t, x) : t \geq T_0\}$ lies in $\overline{U}$, and hence $\omega(x) \subset \overline{U}$. Since $\overline{U} \cap V = \varnothing$ (as $U$ and $V$ are disjoint open sets), we get $\omega(x) \cap V = \varnothing$, contradicting $\omega(x) \cap V \neq \varnothing$. The case $\chi \equiv 1$ is symmetric: it gives $\omega(x) \subset \overline{V}$ and contradicts $\omega(x) \cap U \neq \varnothing$. In both cases we reach a contradiction, so $\omega(x)$ cannot be disconnected. Hence $\omega(x)$ is connected. [/guided] [/step]