Prove that if $A\subseteq B\setminus C$ then A and C are disjoint.

Question

Suppose $x\in A\lor x\in C$, then it can be the case that $x\in A\land x\in C$. However, if $x\in A\land x\in C$, this contradicts $A \subseteq B\setminus C$. Therefore $A\subseteq B\setminus C \rightarrow A\cap C= \varnothing$

Is this proof correct?

Answer 1

It's obvious from the question that if some element $x$ exists in $A$ it will also exist in $B-C$, because of the definition of the subset, now if an element exists in $B-C$, it will exist in $B$ and not exist in $C$, proving that $A$ and $C$ are disjoint as $x \in A \land x \not \in C$

Answer 2

I don't understand why you suppose that $x \in A \lor x \in C$. If $A$ and $C$ are disjoint, then $A \cap C = \emptyset$ and if an element ever belonged to $A \cap C$ one would have $x \in A \cap C \rightarrow x \in A \wedge x \in C$. Thus, one could write the proof as:

Suppose towards contradiction that there is an element $x$ such that $x \in A \cap C$. Then, \begin{align*} x \in A \cap C &\rightarrow x \in A \wedge x \in C\\ &\rightarrow x \in B \backslash C \wedge x \in C\\ &\rightarrow x \in B \wedge x \notin C \wedge x \in C. \end{align*} which is a contradiction derived by the assumption that $A \subseteq B \backslash C$. Thus, there is no element $x \in A \cap C$; i.e., $A \cap C = \emptyset$.

Answer 3

To start with, taking $x\in A\cup C$ is irrelevant and should be omitted, because you take $x\in A\cap C$, which is much stronger a condition.

But the main objection is that you're jumping over the bulk of the proof: you must prove that, if $x\in A\cap C$, then this contradicts $A\subseteq B\setminus C$; you can't just state it as a matter of fact.

Now, where's the contradiction? Since $x\in C$, we know that $x\notin B\setminus C$, by definition of set difference, so a contradiction follows from $x\in A\subseteq B\setminus C$ and $x\in A$.