# $A ⊂ B$ if and only if $A − B = ∅$

I need to prove that $A ⊂ B$ if and only if $A − B = ∅$. I have the following "proof":

$$A \subset B \iff A - B = \emptyset$$ proof for $\implies:$

$$\forall x \in A, x \in B$$

Therefore,

$$A - B = \emptyset$$

proof for $\impliedby$:

If $$A - B = \emptyset$$ then $$\forall x \in B, x \in A$$ since $\forall x \in B, x \in A$,

$$A \subset B$$

However the whole thing seems to be incredibly "fragile" and relies on circular logic (see how I just switched the sets in the for all statements)

Is this a valid proof? Is there a better way to write it?

• The last $\forall x \dots$ should be $\forall x \in A, x \in B$. Mar 1, 2015 at 11:21
• @Bernard Then wouldn't that make my proof the same for both? Mar 1, 2015 at 11:25
• agree with Bernard Mar 1, 2015 at 12:47
• The statement is incorrect. Alberto's answer proves the correct statement (with $\subseteq$ instead of $\subset$), but for some reason doesn't point out this important difference. Mar 24, 2015 at 9:28
• @PeterTaylor "Some authors use the symbols ⊂ and ⊃ to indicate subset and superset respectively; that is, with the same meaning and instead of the symbols, ⊆ and ⊇" en.wikipedia.org/wiki/Subset I personally find it incredibly odd for a very specific and well defined language (math) to be able to use interchangeable symbols like that. My professor uses . Mar 24, 2015 at 10:06

I like using "proof by contradiction" for this one, because it is logically and intuitively clear.

• (⇒) Assume $A\subseteq B$. By contradiction, suppose $A \setminus B \ne \varnothing$. Therefore there exists $x \in A \setminus B$. Therefore $x \in A$ and $x \notin B$ — a contradiction. As a result, $A \setminus B = \varnothing$.

• (⇐) Assume $A \setminus B = \varnothing$. By contradiction, suppose $A \nsubseteq B$. Therefore there exists $x \in A$ such that $x \notin B$. Therefore $x \in A \setminus B$ — a contradiction. As a result, $A \subseteq B$.

As a result, $A \subseteq B$ if and only if $A \setminus B = \varnothing$.   ◻

• Actually, it is a proof by contraposition. Mar 1, 2015 at 11:24
• @Bernard: It is conventionally called proof by contradiction as he correctly states. It's called that because the goal is to derive a contradiction from the additional assumption that the desired conclusion is false. I think you know that.
– MPW
Mar 1, 2015 at 11:29
• I don't see the point, when proving ¬Q -> ¬P, and say ‘suppose P is true and Q false, then deduce P is false’, to call that a proof by contradiction, if you don't need to suppose P true to deduce P false. Mar 1, 2015 at 12:07

One might proceed more directly by noting that $A\subseteq B$ is equivalent to $$\forall x,(x\in A)\implies(x\in B),$$ which is equivalent to $$\forall x,\neg(x\in A)\vee(x\in B).$$ The negation of this is $$\exists x:(x\in A)\wedge\neg(x\in B),$$ which is equivalent to $$\exists x:(x\in A)\wedge(x\notin B),$$ which is equivalent to $$\exists x:(x\in A\setminus B).$$ Renegating then shows us that $A\subseteq B$ is equivalent to $$\forall x,\neg(x\in A\setminus B),$$ at which point we're basically done.

$(\Rightarrow)$; Suppose $A\subset B$. Then $\forall x\in A,x\in B$. Therefore if $x\notin B$ then $x\notin A$. That is if $x\in B^{c}$ then $x\in A^{c}$. Since $A-B=A\cap B^{c}$ we have that $A-B=\varnothing$.
$(\Leftarrow)$; Conversely suppose $A-B=\varnothing$. So $A\cap B^{c}=\varnothing$. Therefore $\forall x\in A,x\notin B^{c}$. Hence $\forall x\in A,x\in B$. So we have that $A\subset B$.
Your proof is valid if one can follow the steps, but the step from $A-B=\emptyset$ to $\forall x\in B, x\in A$ looks a bit hasty and might need some clarification. For example using the definition of $A-B$ being the set of $x: x\in A \land x \notin B$, which in turn is equivalent to $x: x\in B\rightarrow x\in A$ ($\phi\rightarrow\psi$ being the same as $\neg\phi\lor\psi$).
$A - B$ is all elements of $A$ out of $B$.
$A \subseteq B$ means $A$ is surrounded by $B$.
so $\cdots$