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

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 attempt:

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

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

So:

$$A \subseteq \left(A \setminus B \right)$$ (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 $$A \cap B = \emptyset$$

$$A \cap B$$

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

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

$$x \in \emptyset$$

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