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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 $D$.

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    What is $\bigcup \Omega $ supposed to mean? – 2017-02-15
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    Start with a generic element of $\bigcup \Omega$ and argue that if it is in there, it must be in some $A \in \Omega$. – 2017-02-15
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    the union of all the sets in $\Omega$ – 2017-02-15
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    "There could be an element in $\Omega$ not in $A$ that is not in $D$." None of the elements of $\Omega$ are elements of $D$. The elements of $\Omega$ are the sets $A$. $\cup \Omega$ is the set of all elements of elements of $\Omega$, that is, the set of all elements of the $A$s. – 2017-02-15
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    That would be $\bigcup_{A\in \Omega} A$; but OK. – 2017-02-15
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    @zoli Those notations are equivalent – 2017-02-15

3 Answers 3

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Here's a beginning for the proof:

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

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    thanks I will see how far I can take this. – 2017-02-16
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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!

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    I'm finding it extremely difficult to parse that third sentence, and I can't tell what the fourth is referring to. Your sets don't satisfy the if part of the question ($A \subseteq D$ for all $A \in \Omega$) because $\{x\} \not\subseteq D$. – 2017-04-28
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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.