Suppose A $\subseteq$ B. Prove that for every set C, C $\setminus$ B $\subseteq$ C $\setminus$ A.
Seeing that this problem included "every set C", I interpreted the logical form of the goal as follows:
$$\forall C (C\setminus B \subseteq C\setminus A)$$
Expanding $\subseteq$, I got
$$\forall C \forall x (x \in (C\setminus B) \rightarrow x \in (C \setminus A)) $$
To work out a proof, I first supposed some arbitrary set C and then some arbitrary x such that x $\in$ C$\setminus$B. Then, I proved x $\in$ C$\setminus$A.
My question: Do I need to assume some arbitrary set C as the question asks for a proof for every set C?
To me, it seems correct to do so but also awkward. Perhaps this is a consequence of my lacking further insight.
Yes, you have to let $C$ be any arbitrary set, and then verify that $ C \setminus B \subset C \setminus A $
To accomplish this, you can start by picking some $x \in C \setminus B$ (this $x$ is arbitrary, just as the C at the beginning). Now, this means that $x \in C$ but that $x \notin B$. But, by hypothesis, $x \notin A$ then. Therefore $x \in C \setminus A$. In other words,
$$ \boxed{ A \subset B \implies C \setminus B \subset C \setminus A } $$