Generalized Union sets involving Omega

Question

Prove that if $A\subseteq D$ for all $A \in \Omega$ then $\bigcup \Omega \subseteq D$.

Ok so to prove this I was going to start by picking a point in $A$ the obviously its in $D$ and in $\Omega$. But I do not see how this is going to get me that $\bigcup \Omega \subseteq D$.There could be an element in $\Omega$ not in $A$ that is not in $D$.

Answer 1

Here's a beginning for the proof:

Let $x \in \cup \Omega$. Then there exists some $A \in \Omega$ such that $x \in A$.

Answer 2

I'm not sure if you have got the right answer, or the correct way to your question but you should see the next: Suppose $D=\{a,b,c\}$ and $A=\{a\}$ where obviously $A\subset D$. In the other hand, if $\Omega = \{\{a\},\{x\}\} \Rightarrow A\in \Omega$ and $\bigcup \Omega = \{a,x\}$. So, you have an example where $A\subseteq D$ and not for all $A \in \Omega$ then $\bigcup \Omega \subseteq D$ because $x\in \bigcup \Omega$ but $x\notin D$. In other words, it couldn't be true by counter example.

I hope help you with it!

Answer 3

I apologize to my last answer. I've taken with the wrong order your preposition. In order to right it. I post this: $$(\forall A\in \Omega, \quad A\subseteq D) \Rightarrow (\bigcup \Omega \subseteq D)$$ Like our hypothesis. So, it means: $\forall A\in \Omega$, we have $A\in\mathcal{P}(D) \Rightarrow \Omega \subseteq \mathcal{P}(D)$. Moreover: $$ \bigcup\Omega \subseteq \bigcup\mathcal{P}(D)=D \quad \Rightarrow \quad \bigcup\Omega\subseteq D $$ Like we desire.