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

Question

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.

Answer 1

Quite simply: by definition, $C-A\subseteq Cˆ$ and $B-A\subseteq B$. Further, $B\subseteq C$...

Answer 2

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.

Answer 3

Draw Venn Diagrams. This should be obvious.

Now just prove it with Element chasing or definitions or whatever you like.

Answer 4

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 $C$.