So far I have this
Suppose A, B, and C are any set where A $\subseteq$ B.

Proof that A $\cup$ C $\subseteq$ B U C: (Show that x $\in$ B$\cup$C)
Let x $\in$ A $\cup$ C and then by definition of union x $\in$ A or x $\in$ C. So by definition of subset then x $\in$ B. Thus by definition of union x $\in$ B $\cup$ C as was to be shown.

So basically I was wondering do I need two cases for when x is an element of a or b or is this fine? Also I guess with set proof what justifies more than one case?

  • 1
    $\begingroup$ You have an error. ($x\in A$ or $x\in C$) does not imply that $x\in B$. You will need to run two separate cases, the cases being case1) $x\in A$ and case2) $x\in C$. $\endgroup$ – JMoravitz Apr 13 '18 at 4:06
  • $\begingroup$ You could avoid using cases by noting that since $A\subseteq B$ that implies that $B=A\cup (B\setminus A)$. Further, by noting the property that $E\subseteq E\cup F$, we have that $A\cup C\subseteq A\cup C\cup (B\setminus A) = B\cup C$. As for the question of when should or could you use cases, it really depends on the question itself, but usually if there is disjunction going on you have the option of looking at it by cases. $\endgroup$ – JMoravitz Apr 13 '18 at 4:11
  • $\begingroup$ Thanks for the help, your first response is exactly what I was looking for! $\endgroup$ – Nolando Apr 13 '18 at 6:09

Your objective is to prove $A\cup C\subset B\cup C$ for any set $A,B,C$ with $A\subset B$. Now let $x\in A\cup C$ then either $x\in A$ or $x\in C$.

If $x\in A\subset B$ then $x\in B\cup C$.

If $x\in C$ then trivially $x\in B\cup C$.

Hence, $A\cup C\subset B\cup C$.

  • $\begingroup$ Thanks, I think that clears things up a bit! $\endgroup$ – Nolando Apr 13 '18 at 4:23

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