# Non-intersection of sets

I have two 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?

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

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.

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