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The proof I read goes as follows "The statement $\emptyset \subset E$ is equivalent to the statement "If $x \in \emptyset$ then $x \in E$" Since the hypothesis of this if-then statement is false, the implication is true." I understood this as saying that if there is an element in the empty set it is false that it is also in any other set and therefore the empty set is a subset of every set? How does this make any sense, and why does it prove anything?

amWhy
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HakemHa
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    Given a set $S$, the statement "every element of the empty set is in $S$" is vacuously true, since it has no counterexamples. That is, you can not produce an element of the empty set which is not in $S$. – lulu Oct 20 '21 at 19:01
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    If A, then B, is true whenever A is false, no matter what whether B is true or not.. This is the material conditional – amWhy Oct 20 '21 at 19:02
  • It is useful to allow the empty set as a subset of $S$. For instance, if $A,B$ are subsets of $S$, we want to say that their intersection is also a subset of $S$ but this would not always be true if you ruled out $\emptyset$. – lulu Oct 20 '21 at 19:03
  • The hypothesis is not "if there is an element in the empty set". The hypothesis which is always false is "$x$ is an element of the empty set." – Arturo Magidin Oct 20 '21 at 19:34
  • See Truth table for conditional: $\text { False } \to P$ is False for a statement $P$ whatever. Now apply it to "$\text { if } x \in \emptyset \text { then, } P$". – Mauro ALLEGRANZA Oct 21 '21 at 09:14
  • " I understood this as saying that if there is an element in the empty set it is false that it is also in any other set" No. If there is an element in the empty set then it is TRUE that is also in any other set. – fleablood Sep 25 '22 at 06:00

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The statement "If A, then B" is logically equivalent to "A is false OR B is true". In this case, A is the statement "$x\in \emptyset$", which is false. So "A is false OR B is true" is true, hence "If A, then B" is true.

Alternatively, you can examine the contrapositive statement. The contrapositive of "If $x\in\emptyset$ then $x\in E$" is "If $x\notin E$ then $x\notin \emptyset$". The contrapositive statement is true because the conclusion is always satisfied. Hence the original statement is again shown to be true.

Matt E.
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There might be two points that need to be stressed, one axiomatic, one of logical nature:

  • In set theory, the existence of the empty set is derived from the axioms (see the comment by @ArturoMagidin). The statement one derives reads $\exists x \, \forall y \, \neg (y \in x)$ or in English: "There is a set $x$ such that for every $y$, $y$ is not an element of $x$". One can go on and show that there is really only one set with this property, therefore it gets its own symbol $\emptyset$. The takeaway here is where the property $\forall y \, \neg (y \in \emptyset)$ comes from.
  • You mentioned the second point by yourself: it is important to note here that an implication $\phi \to \psi$ is true whenever $\phi$ is false.

To show $A \subseteq B$ for two sets $A$ and $B$, we have to show (by definition) that $\forall y (y \in A \to y \in B)$. For the case in question this means we need to show $\forall y (y \in \emptyset \to y \in B)$. Since $y \in \emptyset$ is false for every $y$ by the defining property of $\emptyset$, every implication is true and therefore also $\forall y (y \in \emptyset \to y \in B)$ as a whole.

Léreau
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  • The empty set's existence does not require a separate axiom. If there is any set at all, then the Axiom of Separation implies the existence of the empty set; and given that the Axiom of Infinity guarantees the existence of a set, that yields the existence of the empty set. – Arturo Magidin Oct 20 '21 at 19:36
  • @ArturoMagidin You are right, thanks for pointing it out! And I only notice now that the refernece I have put does not talk about ZF. I will fix this. – Léreau Oct 20 '21 at 19:39
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If there is an element in the empty set it is false that it is also in any other set and therefore the empty set is a subset of every set? How does this make any sense?

Here is a formal proof using a form of natural deduction (screenshot from my proof checker)

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Dan Christensen
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