0
$\begingroup$

This statement is quite easy to prove using logic ( and the constant F = " Falsum" = the proposition equivalent to any logical falsehood).

But I cannot manage to prove it simply with the laws of the algebra of sets. I mean, without analysing the statement in terms of an arbitrary x belonging to the sets involved.

Is it simply the definition of inclusion in the algebra of sets? ( So that it would not be possible to prove it? )

By the laws of the algebra of sets, I mean :

https://www.math-only-math.com/laws-of-algebra-of-sets.html

$\endgroup$
  • $\begingroup$ What is "Inter Complement"? $\endgroup$ – zoli Apr 14 at 6:58
  • $\begingroup$ @zoll. I meant " the Intersection of the set A and of the set Complement of B" $\endgroup$ – Eleonore Saint James Apr 14 at 7:29
1
$\begingroup$

You are given that $A \subseteq B$, but in your linked laws of algebra for sets, the only principle involving $\subseteq$ is:

$A \subseteq B \Leftrightarrow B' \subseteq A'$

... meaning that there is no rule in this list that allows you to rewrite the $\subseteq$ into any of the set-algebraic operations you really want to work with, i.e. $\cap$, $\cup$, $'$, and $\emptyset$

So no, with the rules as given, you can't derive $A \cap B' = \emptyset$ from $A \subseteq B$

Well, that's no fun!

OK, let's see how we could rewrite the fact that $A \subseteq B$.

Well, one option is to, as you already say, define $A \subseteq B$ as $A \cap B' = \emptyset$ .. but that's no fun either :P

OK, so how about:

$A \cap B = A$

In fact, I really like that one as a definition of $A \subseteq B$: if $A \subseteq B$, then that means that anything in $A$ is automatically in $B$, meaning that intersecting the set with $B$ does not put any further requirements on the objects we're talking about, and so their intersection should indeed just be $A$

So now let's take this piece of information, and see if we can derive $A \cap B' = \emptyset$:

Well:

$A \cap B' \overset{A \cap B = A}{=} (A \cap B) \cap B' \overset{Associative}= A \cap (B \cap B') \overset{Complement}= A \cap \emptyset \overset{Annihilation}= \emptyset$

Yay! (though notice: neither Complement nor Annihilation are in your list of rules ... but they really should be! Those are really, really elementary algebraic principles)

$\endgroup$
1
$\begingroup$

Well, you mean $A\subseteq B\Rightarrow A\cap \overline B=\emptyset$.

[Suppose $x\in A\cap \overline B$. Then $x\in A$ and $x\in \overline B$. Thus by hypothesis $x\in B$ and $x\in\overline B$, i.e., $x\in B\cap\overline B$. But $B\cap \overline B = \emptyset$ and so $A\cap \overline B = \emptyset$. Done.]

We have $A\cap \overline B \subseteq B\cap \overline B$ since by hypothesis $A\subseteq B$ and the intersection operation is monotonous.

Moreover $B\cap \overline B = \emptyset$ and so from above $A\cap \overline B \subseteq \emptyset$. Since also $A\cap \overline B\supseteq \emptyset$, we obtain $A\cap \overline B=\emptyset$. Done.

$\endgroup$
  • 1
    $\begingroup$ The OP asked for the way to prove it without using this method. $\endgroup$ – Graham Kemp Apr 14 at 7:47
  • $\begingroup$ My question was to know whether the proposition could be derived from some ( hypothetically) more primitive formula, without using any set theoretic tool ( such as the membership relation). In other words, whether it could be proved inside the algebra of sets as a self contained theory). $\endgroup$ – Eleonore Saint James Apr 14 at 7:49

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.