Prove: if $\Omega \subseteq \Gamma$, then $\bigcup\Omega \subseteq \bigcup\Gamma$
I know that $\bigcup\Omega$ is the set of all the elements in $\Omega$ (not a precise definition) but I am unsure on how to use this information to complete the proof.
My initial thought is to let $x \in \Omega$ then we know $x \in \Omega, \bigcup\Omega$ does that imply $x \in \bigcup\Gamma$?
On
$\bigcup \Omega~$, the union of all set-elements of $\Omega$, is shorthand for $~\bigcup\limits_{X\in \Omega} X$
We argue that if $\Omega\subseteq \Gamma$ then every set in $\Omega$ is in $\Gamma$, so their union is a subset of the union of every set in $\Gamma$.
Alternatively: If $\Omega\subseteq\Gamma$, then and only then $\Gamma = \Omega\cup(\Gamma\setminus\Omega)$. If so then $\bigcup \Gamma = \bigcup\Omega\cup\bigcup(\Gamma\setminus\Omega)$. If so, therefore $\bigcup\Omega\subseteq \bigcup\Gamma$.
$$\begin{align}\bigcup_{X\in\Gamma} X ~&=~ \bigcup_{X\in\Omega\cup(\Gamma\setminus\Omega)} X\\ ~&=~ \bigcup_{X\in\Omega}X \cup\bigcup_{X\in(\Gamma\setminus\Omega)}X \\[2ex] \bigcup\Gamma ~&=~ \bigcup\Omega\cap\bigcup(\Gamma\setminus\Omega)\end{align}$$
On
Definition: let be $A$ a set: $$ \bigcup A:=\{x|\exists z\in A:x\in z\}$$
Thereom: let be $\Omega$, $\Gamma$ sets: $$ \Omega\subseteq \Gamma \to \bigcup \Omega \subseteq \bigcup\Gamma$$
Proof: $$\begin{align} x \in \bigcup \Omega &\leftrightarrow \exists z \in \Omega: x \in z \\ &\to \exists r \in \Gamma: x \in r \text{ (namely } r=z \text{, and } z \in \Gamma \text{ because } \Omega \subseteq \Gamma) \\ &\to x \in \bigcup \Gamma \text{ (by definiton)} \end{align}$$
$\bigcup \Omega$ is the union of all elements in $\Omega$. For example if $ \Omega = \{ \omega_1, \omega_2, \dots\}$ then $\bigcup \Omega = \omega_1 \cup \omega_2 \cup \dots$
More formally $ \forall X \in \Omega . x \in X \Rightarrow x \in \bigcup \Omega $. To prove your implication can can chase elements from the sets that are elements of the $\Omega$ omega set into the $\Gamma$ set.