Set proof for intro to pure math class: $(A-C)\subseteq (A-B)\cup (B-C)$

Question

I don't know if this question has been asked but I couldn't find it. The question is: Suppose $A,B,C$ are sets. Prove that $(A-C)\subseteq (A-B)\cup (B-C)$.

What I have is: Let $x\in (A-B)\cup (B-C)$ then $x\in A$ and $x\notin B$ or $x\in B$ and $x\notin C$. Hence, $x\in (A-C)\implies (A-C)\subseteq (A-B)\cup (B-C)$.

Now my question is, is it acceptable to work from the right to the left as I have just done. My conclusion follows intuition; however, it seems a bit unformal to me. Thanks!

---

EDIT

For my second attempt, I broke the problem into two, when $x\in B$ and $x\notin B$.

Let $x\in (A-C)$ then $x\in A$ and $x\notin C$. Now, if $x\in B$ then, $x\notin (A-B)$ but $x\in (B-C)$. So, $x\in (A-B)\cup (B-C)$. If $x\notin B$, then $x\in (A-B)$ and $x\notin (B-C)$. So, $x\in (A-B)\cup (B-C)$.

Is my proof valid now?

Answer 1

Let $x\in (A-C)$ then $x\in A$ and $x\notin C$. Now, if $x\in B$ then, $x\notin (A-B)$ but $x\in (B-C)$. So, $x\in (A-B)\cup (B-C)$. If $x\notin B$, then $x\in (A-B)$ and $x\notin (B-C)$. So, $x\in (A-B)\cup (B-C)$.

Yes, that is the form for a proof by cases, and the reasoning in each branch is valid. It may need some polish, but that's the shape of it.

For instance, you need to complete it. "Then in either case any $x\in A\smallsetminus C$ infers that $x\in(A\smallsetminus B)\cup (B\smallsetminus C)$, so therefore $A\smallsetminus C \subseteq (A\smallsetminus B)\cup (B\smallsetminus C)$." or such.

Answer 2

Your proof does not imply $$(A-C)\subseteq (A-B)\cup (B-C)$$

Note that

If $x$ is in $A-C$ the $x$ is in $A$ and $x$ is not in $C.$

There are two cases regarding $x$ and $B.$

If $x$ is in $B$ it is in $B-C$

If $x$ is not in $B$, then it is in $A-B$

In either case $x$ is in the union $ (A-B)\cup (B-C)$

That is $$(A-C)\subseteq (A-B)\cup (B-C)$$

Answer 3

When you union $(A-B)$ with $(B-C)$ you are "putting back" into $A$ all the elements in $B$ that you took away except for those that are in $C$. $A$ less all the elements in $C$ must be equal to or a subset of this.