I have two sets, $A$ and $B$: Sets A and B.

$B$ is in $A$, so $A \cap B = B$.

I would like to mathematically express the non-intersection of these sets. It is not the same $C= A \setminus B$ or $B \setminus A$ than $A$ (non $\cap$) $B$.

In this case it will be $A \setminus B$, but how I now if $A\subset B$ or $B\subset A$? How can I do it?

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    $\begingroup$ Non intersection of $A$ and $B$ ? $A \cap B = \emptyset$. $\endgroup$ – Mauro ALLEGRANZA May 29 '19 at 13:44
  • $\begingroup$ I mean, the subset of elements that are not in the other. $\endgroup$ – Sergio Cavero May 29 '19 at 13:46
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    $\begingroup$ Are you referring to the symmetric difference? en.wikipedia.org/wiki/Symmetric_difference $\endgroup$ – tia May 29 '19 at 13:49
  • $\begingroup$ Would that not be $A'$? Since $B\subset A$. The elements not in either would just be the elements not in $A$. $\endgroup$ – Tom Himler May 29 '19 at 13:49
  • $\begingroup$ @tia that's what I was looking for! I have never heard about it in my short life. Thanks $\endgroup$ – Sergio Cavero May 29 '19 at 13:53

$B\subseteq A$ is equivalent with $B\setminus A=\varnothing$.

So if that is the case then for the symmetric difference we find: $$A\Delta B=(A\setminus B)\cup (B\setminus A)=(A\setminus B)\cup \varnothing=A\setminus B$$

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First a remark :

It is better not to use the expression “ negation” when one talks about a set; a set cannot be negated, only a sentence can be negated; but one can say that the set called “complement of the set A ∩ B” is defined by the negation of the sentence defining A ∩ B. That is, if the symbol for the complement of A ∩ B is : (A ∩ B)’ then one can say that :

(A ∩ B)’ = the set of all x such that it is false that (x belongs to A & x belongs to B).

The complement of a set S , denotetd by the symbol : S' , is the set of all x ( belonging by definition to the universal set U) that do not belong to B, that is the set : U – S.

Now regarding your question :

If the set A also is your universal set U , that is, if U=A

then , in that special case :

A-B = U-B = the set of all x that do not belong to B= B’ = complement of B

(A ∩ B)’ = U – (A ∩ B) = U – (U∩ B) = U – B = B’ = complement of B.

and therefore ( still in the special case we are considering) :

(A ∩ B)’ = A – B.

But that is not true in general. If A is not the the universal set U ( and , in general, when one talks about two sets A and B, they are supposed to be different from the universal set) :

(A ∩ B)’ ≠ A – B.

As to your last question, the hypothesis of your problem does not allow you to determine whether A is or is not also included in B. Knowing that B is included in A does not rule out the possibility A to be also included in B, but it does not imply this either.

In case (a) neither A nor B is empty, (b) neither A nor B is the universal set and (b) the two sets are included one in the other , that is, in case they are equal , then :

(A ∩ B)’= (A∩ A)’ = A’ = complement of A = U – A.

A – B = A – A = ∅

So in that case : (A ∩ B)’ ≠ A – B

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