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According to definition of subset, if A is subset of B, it means every element in A is also in B, but how can apply this logic in case of null set?

Asaf Karagila
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Shivam Maurya 1729
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    $\emptyset \subset B$ would mean that every element of $\emptyset$ is an element of $B$. Is there any element of the null set that isn't in set B? How many elements of $\emptyset$ can you name? Can you name any that are not in $B$? Is it true or is it not true that every single element of $\emptyset$ that you can name and exist (all zero of them) are elements of $B$? As there are no elements of $\emptyset$ at all then I can't name any that are not in $B$ and every one I can name (I can't name any) are in $B$ (as I can't name any). – fleablood Sep 25 '22 at 05:56
  • See also https://math.stackexchange.com/a/48202/ – GEdgar Sep 25 '22 at 07:10
  • The statement $A\subset B$ is either true or false. If it is false then there has to be an element in $A$ that is not in $B$. Since no such element can exist for the empty set, the statement that the empty set is a subset of every other set must be true. – John Douma Sep 25 '22 at 07:19
  • @JohnDouma The law of excluded middle is useless here. – Anne Bauval Sep 25 '22 at 07:40
  • @AnneBauval I would love to see a proof of this without it. The accepted answer in your link uses the fact that $A\implies B$ is equivalent to $\lnot A\lor B$. That requires the excluded middle to prove. – John Douma Sep 25 '22 at 13:53
  • @JohnDouma a proof without it is the one by Léreau (in my link). It does not use the excluded middle but only the fact that (even in intutionistic logie) $(\neg \phi \vee \psi) \to (\phi \to \psi)$. – Anne Bauval Sep 25 '22 at 15:24
  • @AnneBauval $A\to A$ is the same as $\lnot A\lor A$, the excluded middle. I agree that $\lnot A\lor B\to (A\to B)$ without the excluded middle, but why does it follow that either an element is not in the empty set or in every other set?. Also, the proof you allude to states that $A\to B$ is always true when $A$ is false. That follows from $(A\to B)\to (\lnot A\lor B)$. That requires the excluded middle. – John Douma Sep 25 '22 at 16:39
  • @JohnDouma The indicated proof does not care about $A\to A$ and does not need any "either...or...". That $A→B$ is always true when $A$ is false does not follow from $(A\to B)\to (\lnot A\lor B)$ but from the converse, which, as said before, does not require the excluded middle. – Anne Bauval Sep 25 '22 at 16:52

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