Rigorously prove: If $A\setminus B = A$, then $A \cap B = \emptyset$.

Logically it makes sense i.e. since if $A\setminus B$ is equal to $A$, then $B$ must be the empty set hence $A \cap B$ must be equal to the empty set since anything intersected with the empty set is the empty set.

My main question is how I can go about this to prove it formally, i.e 'mathematically/rigorously'

My attempt:

$\left(A \setminus B \right) \subseteq A$ (1)

$A$ is a subset of $A$ hence also a subset of $\left(A \setminus B \right)$


$A \subseteq \left(A \setminus B \right)$ (2)

Combining (1) and (2) we are allowed to conclude that:

$ \left(A \setminus B \right) = A$

Hence since $ \left(A \setminus B \right) = A$, then $B = \emptyset$ then rtp that $A \cap B = \emptyset$

$A \cap B$

$(x \in A \land x \in B)$

$(x \in A \land x \in \emptyset)$

$x \in \emptyset$

Thanks in Advance

  • $\begingroup$ Welcome to Maths SX! You mean $A\smallsetminus B$, not $A/B$, I suppose? $\endgroup$
    – Bernard
    Feb 1 '19 at 12:29
  • $\begingroup$ Apparently so (according to everyone on this forum) ... however this is how it is written on the textbook I'm using $\endgroup$
    – xalalau
    Feb 1 '19 at 12:41
  • $\begingroup$ What textbook is that? $\endgroup$
    – bof
    Feb 1 '19 at 12:42
  • $\begingroup$ Note that $A \cap B$ and $A \setminus B$ are two disjoint subsets of $A$. $\endgroup$ Feb 1 '19 at 12:44

" Hence since $A\setminus B = A$, then $B = \varnothing$ then rtp that $A\cap B=\varnothing$"

Nowhere we reach the conclusion that $B=\varnothing$.

This is a way to do it:

Assume that $x\in A\cap B$.

Then $x\in A=A\setminus B$ and $x\in B$.

Then $x\notin B$ and $x\in B$ so a contradiction is found.

We conclude that the assumption $x\in A\cap B$ is false, and this for every $x$.

That means exactly that $A\cap B$ is empty.


Note that it is always true that

$$A \setminus B = A \cap B^C$$

Hence, if $A \setminus B =A$, then $A \cap B^C =A$, and so:

$$A\cap B= (A \cap B^C) \cap B= A \cap (B^C \cap B)=A \cap \emptyset=\emptyset$$


A correct proof is sketched in the other answer. To answer your question about whether your reasoning is correct: Not really. You're trying to show that if $A\setminus B=A$ then $\cap B=\emptyset$. So $A\setminus B=A$ is given. Hence a line saying "we may conclude $A\setminus B=A$ makes very little sense.

Not that it matters, since you're given that $A\setminus B=A$, but your proof of this fact is also wrong. You state that $A$ is a subset of $A\setminus B$, but that's not true in general.


Suppose $x\in A\cap B$. Then $x\in A$ and $x\in B$. But then $x\not\in A\setminus B$. Since $A\setminus B=A$, this leads to a contradiction. Done.


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