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?


  • 1
    $\begingroup$ "Suppose $x \in A\backslash B$" is the only change I'd suggest. Overall a nice proof. $\endgroup$ – MerylStreep Sep 13 '15 at 19:27
  • $\begingroup$ Oh cool! Thanks! $\endgroup$ – Mathy Sep 13 '15 at 19:31
  • $\begingroup$ Welcome to our site! $\endgroup$ – 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)

  • $\begingroup$ Thanks. In what others ways can we solve/write the proof? apart from this way? $\endgroup$ – Mathy Sep 13 '15 at 19:36
  • $\begingroup$ 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. $\endgroup$ – Ivo Terek Sep 13 '15 at 19:40
  • $\begingroup$ ah, i see! Thanks anyways (: $\endgroup$ – Mathy Sep 13 '15 at 19:41

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