2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
1
$\begingroup$

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

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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.