How can I prove that for any $A, B$ if $A\subseteq B$ and $B\subseteq C$, then $(C-A)\cup (B-A)\subseteq C$?

How can I prove that for any $$A, B$$ if $$A\subseteq B$$ and $$B\subseteq C$$, then $$(C-A)\cup (B-A)\subseteq C$$?

I've been working on this question and I haven't really made much progress with it. I know that I can rewrite it as $$(C \cap A^c) \cup (B\cap A^c)$$. I'm pretty sure that if $$A \subseteq B$$ and $$B \subseteq C$$ then $$A \subseteq C$$. If this is the case then wouldn't $$(C \cap A^c) = \emptyset$$ and $$(B \cap A^c) = \emptyset$$ or am not understanding something with set theory? Thank you for the help.

• Rewrite each statement as an implication in memberships. – J.G. Dec 9 '18 at 22:38
• Consider $A=\{1\}, B=\{1,2\}, C=\{1,2,3\}$. We have $A\subseteq B\subseteq C$. But the complement $A^c$ of $A$ with respect to $C$ (as it's important to stress what the complement is in respect to) is $\{2,3\}$, so $C\cap A^c=\{2,3\}$ and $B\cap A^c=\{2\}$. – Shaun Dec 9 '18 at 22:39
• $A \subset C$ means everything in $A$ is in $C$. $C\setminus A$ means all the stuff in $C$ that isn't in $A$. If $A$ is in $C$ then $C\setminus A$ is all the stuff in $C$ that isn't in $A$. That needn't be empty. Maybe you are confusing $C \setminus A$ with $A \setminus C$? Setminus is not commutative. $A \setminus C$ is all the stuff in $A$ that is not in $C$, but everything in $A$ is in $C$ so $A\setminus C = \emptyset$. But $C\setminus A$ is just... $C \setminus$A\$. – fleablood Dec 9 '18 at 23:09

Quite simply: by definition, $$C-A\subseteq Cˆ$$ and $$B-A\subseteq B$$. Further, $$B\subseteq C$$...
It's not the case that $$X \subseteq Y$$ implies $$Y-X = \emptyset$$. Indeed if $$X = \{0\}$$ and $$Y = \{0,1\}$$ then $$Y - X = \{1\}$$.
We do have, however, that $$X \subseteq Y$$ implies $$Y - X \subseteq Y$$. Applying this to what you have so far, conclude that $$C- A \subseteq C$$ and $$B - A \subseteq B \subseteq C$$. Hence $$(C-A) \cup (B-A) \subseteq C$$ too.
For every $$x\in (B-A)\cup (C-A)$$, either $$x\in B-A$$ or $$x\in C-A$$ (or both). For every $$x\in B-A$$, $$x\in B$$ hence $$x\in C$$. For every $$x\in C-A$$, obviously $$x\in C$$. Hence if $$x\in (B-A)\cup (C-A)$$, surely $$x\in C$$. By definition, this means that it is a subset of $$C$$.