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Prove if $A$ is a subset of $B$, then $A\backslash B= \emptyset$.

I used a proof by contrapostive.

The assumption ($A\backslash B$ not being equal to the empty set) means there exists an $x$ such that $x\in A\backslash B$ implies $x\in A$ and $x\not\in B$. This implies that $A$ is not a subset of $B$.

How is this proof, thanks.

Carlos Bacca
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    Your proof is fine in terms of content, but in terms of presentation you have mixed notations. You wrote both $A/B$ and $A\setminus B$ to denote the set difference. Only the second should be used. Writing by hand I'd expect you would have access to whatever symbols you want, but if typing here you can find several symbols here in this MathJax and $\LaTeX$ primer. In particular \not\subset and \not\in – JMoravitz Nov 07 '18 at 17:27
  • Thanks ill learn to format – Carlos Bacca Nov 07 '18 at 17:29
  • "How is this proof" I'm not sure what this means. Are you asking if that is a valid proof? Yes, it is valid. Or are you asking how does that prove $A \not \subset B$? Or are you asking how does that prove $A\setminus B = \emptyset$? – fleablood Nov 07 '18 at 17:30
  • i was just asking if it was valid – Carlos Bacca Nov 07 '18 at 18:22

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Your proof is correct. You could have avoided using the contrapositive and just proved it directly:

Suppose that $A \subseteq B$. If there is some $x \in A \setminus B$, then $x \in A$ and $x \not\in B$, which contradicts the assumption that $A \subseteq B$; hence $A \setminus B$ has no elements, so is empty.

It doesn't really make much of a difference to the structure of the proof, though—the original statement is of the form $p \Rightarrow \neg q$, so its contrapositive is also of the form $q \Rightarrow \neg p$ (where $p$ means '$A \subseteq B$' and $q$ means '$A \setminus B$ is inhabited').

But as a general rule, it's nice not to use indirect proof techniques if you don't have to.

Clive Newstead
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