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Show that $A\subseteq B$ if and only if for any set $C$, one has $(A\cup C)\subseteq (B\cup C)$

$\exists x\in A\cup C \land \exists x\in B\cup C$

$(\exists x\in A\lor \exists x\in C) \land (\exists x\in B\lor \exists x\in C)$

$(\exists x\in A\land \exists x\in B) \lor (\exists x\in C)$

How do i show that $(\forall x\in A\land \exists x\in B) \equiv A\subseteq B$ can i infer this is true?

Then i'm stuck. Or is my argument even correct for that better?

Crostul
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Danxe
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1 Answers1

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Just handle each side in turn. First, assume $A\subseteq B$ Let $C$ be an arbitrary set. To show one set is a subset of another, we assume the existence of an element of the first, and show that means that element has to be in the second. (note if there are no actual elements, this still works fine, since the empty set is a subset of anything).

So, let $x\in A\cup C$ be arbitrary. Then $x\in A \vee x\in C$ Taking in cases, if $x \in A$, then since $A\subseteq B$, $x\in B$, hence $x\in B \cup C$. Otherwise, $x\in C$ implies $x\in B \cup C$. Thus, in all cases we have $x\in A\cup C\implies x\in B\cup C$, thus $A\cup C\subseteq B\cup C$.

The reverse direction is even easier. Assume $A\cup C\subseteq B\cup C$ for each C. Now just let $C=\emptyset$, and you get $A\cup \emptyset=A\subseteq B\cup \emptyset=B$, and you're done.

Alan
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  • How do you ensure $A\cup C \subseteq B\cup C \Rightarrow A \cup B$ just by proving the null set? – Danxe Sep 06 '15 at 09:25
  • You're assuming the right hand side works for ANY C. Thus, you can pick one and it works, so pick the empty set. – Alan Sep 06 '15 at 09:26
  • But isn't it true that proving the evidence of a particular truth doesn't actually prove the evidence of truth? – Danxe Sep 06 '15 at 09:29
  • I'm not following your question. – Alan Sep 06 '15 at 09:55