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Proof 1:

Let $S$ et $E$ be 2 sets.

If $E$ isn't a subset of $S$, then we have $\exists x: x\in E \not\rightarrow x\in S$, which implies that E is not an empty set.

Take the contrapositive of "if $E$ isn't a subset of $S$, then $E$ is not an empty set", and we have:

"if $E$ is an empty set, then $E$ is a subset of $S$".

Proof 2 (by contradiction):

Let $S$ be a set, and $E$ an empty set. By the definition of an empty set, we know that $\forall x: x\notin E$.

Suppose $S$ is a set that does not have $E$ as its subset. If $E$ isn't a subset of $S$, then we have $\exists x: x\in E \not\rightarrow x\in S$, which implies that E is not an empty set. This leads to a contradiction, therefore $S$ cannot exist. We conclude that the empty set is a subset for all sets.

Heng Wei
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  • Duplicate here – Arctic Char Jul 10 '22 at 23:31
  • The standard proof/argument is that every statement about a non-existent object is vacuously true. As $E$ has no elements, any statement about non-existent elements of $E$ is vacuously true. So saying they are all elements of $S$ is true. This drives novice students nuts. I like your way because it shifts the burden to something direct and non-vacuous. ... I wonder though about what pure logicians would make of the statement "If $A$ is not a subset of $B$ there exists an element of $A$ that is not an element of $B$". I say it's acceptable. At any rate that gets into more theory than needed – fleablood Jul 10 '22 at 23:37
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    Does this answer your question? Prove that the null set is a subset of a set $A$. – Ethan Bolker Jul 10 '22 at 23:38
  • Oh.... and yes, your proof is perfectly valid. (Unless someone has issue with the excluded middle and proofs by contradiction.) – fleablood Jul 10 '22 at 23:39
  • @fleablood If you define universal quantification as "shorthand" for the negation of existential quantification you can make that idea explicit. Frequently that's done as when proving theorems in logic its easiest to have as few formal symbols as possible. For example, if A then B is treated as shorthand for "not A or B" and conjunction is defined in terms of negation and disjunction using DeMorgan's Laws. – David Reed Jul 10 '22 at 23:42
  • @DavidReed how can I explicitly create a contradiction? – Heng Wei Jul 10 '22 at 23:42
  • @DavidReed is it now explicitly expressed? – Heng Wei Jul 10 '22 at 23:50
  • What you have proven is "If $E$ is not a subset of $S$ then $E$ is not empty". I disagree with David Reed in that as you had stated $E$ was empty, concluding $E$ is not empty is indeed a contradiction.... However as you never actually used the fact that $E$ was empty this works better as a contrapositive proof. You have proven "If $E$ is not a subset of $S$ then $E$ is not empty" and that is equivalent to the contrapositive, is if $E$ is empty there is no so set that it is not a subset of and therefore it is an emptyset is a subset of all sets. – fleablood Jul 11 '22 at 00:07
  • @fleablood 'I wonder though about what pure logicians would make of the statement "If $A$ is not a subset of $B$ there exists an element of $A$ that is not an element of $B$".' Why would logicians have an issue with this statement? – Jair Taylor Jul 11 '22 at 00:23
  • No need for contradiction. You've shown that $\emptyset$ cannot fail to be a subset of any set, hence it is a subset of every set. – Charles Hudgins Jul 11 '22 at 00:33
  • I would actually still clean it up a little. For starters, you want to assume the hypothesis is false. You haven't made that assumption explicit. So you want to start by assuming there exists a set S such that the empty set is not a subset of S, and get rid of "if E is a subset of S part, that is not needed", and show how the latter statement implies E is not empty. – David Reed Jul 11 '22 at 00:33
  • @DavidReed I added another proof, do you think now the contradiction and the assumption are explicit? – Heng Wei Jul 11 '22 at 01:06
  • I posted an answer. Not ideal, but if nobody else helps I will see if I can link a handwritten picture.This also depends on how strict your professor is. If they are willing to accept that "there exists an x such that not(if x is in E then x is in S)" as implying "there exists an x in E" without any further justification, then what you had initially was correct. – David Reed Jul 11 '22 at 01:20

2 Answers2

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I don't remember Latex very well, and that is whats frustrating to me, as its difficult to explain this non-symbolically. You were on the right track initially. You just needed to show that "there exists an x such that not(if x is in E then x is in S" implies "there exists an x such that x is in E".

"If A then B" is equivalent to "not(A) or B". So a negated conditional is equivalent to "not(not(a) or B), which by demorgan's law is equivalent to "A and not B".

So, letting P(x) be "x is an element of E" and Q(x) be"x is an element of S"

"There exists an x, such that not(if P(x) then Q(x))" is equivalent to "there exists an x such that (P(x) and not(Q(x)). Which implies "there exists an x such that P(x)." However, the definition of the empty set is "for all x, not P(x). That is the contradiction. You basically are showing that the empty set is not empty.

You just needed to carry that last part out a bit further.

David Reed
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I think this is in general a pretty slippery statement to prove, so here are just some attempts at proving it clearly. I'm still not totally satisfied with how I've stated things, but I've never read a proof of this fact that doesn't feel like it's hiding something.

I would write your proof like this: Fix a set $S$. A set $E$ is not a subset of $S$ if and only if there exists $x \in E$ such that $x \not\in S$. There exists no $x \in \emptyset$, so there certainly does not exist $x \in \emptyset$ such that $x \not\in S$. We conclude that $\emptyset$ cannot fail to be a subset of $S$. In other words (if we accept double negation) $\emptyset \subset S$.

We could also have a proof by contraposition. By definition, $E \subset S$ if and only if for all $x$, $x \in E$ implies $x \in S$. Taking the contrapositive, $E \subset S$ if and only if for all $x$, $x\not\in S$ implies $x \not\in E$. For all $x$, $x \not\in \emptyset$, so $x \not\in S$ trivially implies $x\not\in \emptyset$.

Charles Hudgins
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