# Proof writing: A\B subset A

Proof: A\B $\subset$ A

Suppose A\B.

This means x $\in$ A and x $\notin$ B. Then x $\in$ A.

Therefore, A\B $\subset$ A.

I know this is a fairly easy problem. I'm just trying to learn how to write proofs the proper & correct way. My question, is this correct way to write the proof? what am I missing? Any tips to improve it?

Thanks.

• "Suppose $x \in A\backslash B$" is the only change I'd suggest. Overall a nice proof. – MerylStreep Sep 13 '15 at 19:27
• Oh cool! Thanks! – Mathy Sep 13 '15 at 19:31
• Welcome to our site! – kjetil b halvorsen Sep 13 '15 at 19:33

Your proof is fine. What you essentialy did was: $$x \in A \setminus B \implies x \in A,$$which means that $A \setminus B \subseteq A$. As pointed in a comment, the correct would be to start with "Suppose $x \in A \setminus B$" ------- "then $x \in A$". But your idea is correct. (As a side note, you can use \setminus to produce $\setminus$, \in to produce $\in$, and \subset to produce $\subset$, in case you're not used to $\TeX$ yet)
• To be honest, I don't know. You did what it had to be done. If we had $A,B \subset U$ ($U$ would be a universe set), then we could say $A \setminus B = A \cap B^c \subseteq A$, where $B^c$ is the complement of $B$ w.r.t. $U$ and we use a property for $\cap$. But I'm stretching it here, keep it simple. – Ivo Terek Sep 13 '15 at 19:40