# Is my proof valid for: If $A-B \subseteq C$ and $A \not \subseteq C,$ then $A \cap B \not = \emptyset$.

Just starting to learn proofs and could use some feedback:

Statement: Let A, B, and C be nonempty sets. If $A-B \subseteq C$ and $A \not \subseteq C,$ then $A \cap B \not = \emptyset$.

Proof: Since $A \not \subseteq C$, there exists an $x \in A$ such that $x \not \in C.$

Since $x \not \in C$ and $A - B \subseteq C$, $x \not \in A - B$. Since $x \in A$ and $x \not \in A - B$, $x \in B$.

Hence $x \in A \cap B$. That is, $A \cap B \not = \emptyset.$

Thank you.

Idle Math Guy

• looks good to me – Andreas Blass Sep 29 '16 at 18:44
• It's OK, but it much simpler to use the contradiction: if $A\cap B=\varnothing$ then $A-B=A$, and immediate contradiction. – Canis Lupus Sep 29 '16 at 18:48
• It's important to note that $-$ is the complement, and does not refer to the Minkowski difference. – LinAlg Sep 29 '16 at 18:58
• A - B is set difference. i.e. All the elements in the set A that are not in the set B. ${ x | (x \in A) \land (x \not \in B)}$. – Idle Math Guy Sep 29 '16 at 19:42
• @Idle Math Guy:Was your original question answered? If so, please, provide an answer and so mark your question as settled. – Moritz Oct 1 '16 at 15:32

## 1 Answer

Proof is a valid proof. I do like the contradiction route that Corvus used...I think it is more concise.