Omega limit-set is union of omega limit points

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Given some dynamical system $(X, \mathbb{T}, S_t)$. Prove for any $B \subset X$ the inclusion holds; $$\bigcup_{b \in B} \omega(B) \subset \omega(B)$$

From my understanding, it could be equivalent to prove that $\omega(\omega(B)) \subset \omega(B)$ which is obvious but difficult to formulate as a proof.

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Here, for a set $C \subset X$, we have $$ \omega(C) = \{ \text{limit points of the form } \lim_{n \to \infty} S_{t_n} b_n \, , \text{ where } t_n \to \infty, \{ b_n \} \subset C\} \, . $$

To prove $$ \bigcup_{b \in B} \omega(b) \subset \omega(B) \, , $$ for any $B \subset X$, let $x \in \omega(b)$ for some $b \in B$. Then, $x = \lim_{n \to \infty} S_{t_n} (b)$ for some sequence $t_n \to \infty$. But notice that if we define the sequence $\{b_n \} \subset B$ by setting $b_n = b$ for all $n$, then $x$ is a limit point of the desired form $x = \lim_{n \to \infty} S_{t_n} b_n$. Thus $x \in \omega(B)$.