for all sets A, B, and C if A ⊆ B then A$\cup$C ⊆ B$\cup$C.

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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?

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