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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 $x \in \bigcup\Gamma$?

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    Wouldn't "the set of all the elements in $\Omega$" simply be $\Omega$? The only set union notation I'm familiar with always involves more than one set, so the meaning of $\cup \Omega$ is unclear to me. – 2017-02-14
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    My interpretation is that $\Omega$ could have sets with in itself and the union of all of those "reduces" to only the unique elements. @FalafelPita – 2017-02-14
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    The idea in set theory is that .... everything .... is a set. So, $\bigcup \Omega$ is just the union of all the elements of the set $\Omega$, **each** of which is a set. – 2017-02-14
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    ECollins: It is important in cases like this to be absolutely precise. $\bigcup\Omega$ is _not_ the set of all elements in $\Omega$; @FalafelPita is correct that the set of all elements in $\Omega$ is simply $\Omega$. $\bigcup\Omega$ is the _union_ of all the elements in $\Omega$ (note that the elements of $\Omega$ are themselves sets). So for instance if $\Omega=\{\{1,2\},\{2,3,4\}\}$, then $\bigcup\Omega = \{1,2,3,4\}$. This is not equal to $\Omega$. – 2017-02-14

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$\bigcup \Omega~$, 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 $\bigcup\Omega\subseteq \bigcup\Gamma$.

$$\begin{align}\bigcup_{X\in\Gamma} X ~&=~ \bigcup_{X\in\Omega\cup(\Gamma\setminus\Omega)} X\\ ~&=~ \bigcup_{X\in\Omega}X \cup\bigcup_{X\in(\Gamma\setminus\Omega)}X \\[2ex] \bigcup\Gamma ~&=~ \bigcup\Omega\cap\bigcup(\Gamma\setminus\Omega)\end{align}$$

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$\bigcup \Omega$ 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.

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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: $$\begin{align} x \in \bigcup \Omega &\leftrightarrow \exists z \in \Omega: x \in z \\ &\to \exists r \in \Gamma: x \in r \text{ (namely } r=z \text{, and } z \in \Gamma \text{ because } \Omega \subseteq \Gamma) \\ &\to x \in \bigcup \Gamma \text{ (by definiton)} \end{align}$$