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This question already has an answer here:

$\hskip8pt$ Definition. If $A$ and $B$ are sets, then $A$ is a subset of $B$ iff every element of $A$ is also an element of $B$.

The empty set $\{\}$ is a subset of every set because, if $A$ is an arbitrary set, then every element of $\{\}$ is also an element of $A$. This statement is said to be "vacuously true" because $\{\}$ has no elements.

But what about statements that imply the negation of this?

Namely: the empty set $\{\}$ is not a subset of any set because, if $A$ is an arbitrary set, then not every element of $\{\}$ is an element of $A$. Couldn't this statement be vacuously true because $\{\}$ has no elements?

What about: the empty set $\{\}$ is not a subset of some sets because there exists a set $A$ with the property that not every element of $\{\}$ is an element of $A$. Couldn't this statement be vacuously true because $\{\}$ has no elements?

I understand that the fact that the empty set is a subset of every set (by vacuous truth) is merely a convention (I might be wrong, and it might be a necessity, to avoid contradictions or something). Is it possible to develop a (contradiction-free) mathematical formalism where we adopt the opposite convention? If so, what are the consequences?

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marked as duplicate by Andrés E. Caicedo, Asaf Karagila elementary-set-theory Jan 2 '18 at 19:38

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  • $\begingroup$ Every element of the empty set is in $A$, and every element of the empty set is not in $A$. These statements don't contradict each other. $\endgroup$ – Matthew Leingang Jan 2 '18 at 17:41
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    $\begingroup$ @MatthewLeingang: The O.P. wrote ‘not every element of {} is in $A$’, which is not the same as ‘every element of {} is not in $A$’ $\endgroup$ – Bernard Jan 2 '18 at 17:45
  • $\begingroup$ @Bernard: you're right about that. Although I'm not sure how to get closer to answering the question. $\endgroup$ – Matthew Leingang Jan 2 '18 at 17:49
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    $\begingroup$ As soon as you have a statement which requires the existence of an element of the empty set you have to be careful, because no such element exists. Note that "not every" means "there exists a counterexample" $\endgroup$ – Mark Bennet Jan 2 '18 at 18:13
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    $\begingroup$ While one could argue that this is not a duplicate to the question I offered, since the question here is based on the false premise that the empty set is only taken as a subset of every set as a convention, any good answer would invariably be an answer explaining it is not a convention, but a theorem. $\endgroup$ – Asaf Karagila Jan 2 '18 at 19:41
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The fact that the empty set is a subset of every set is a direct consequence of the definitions of the terms "subset" and "empty set".

If a set $B$ is not a subset of a set $A$, then by the definition of "subset" this means that there is an element of $B$ that is not an element of $A$. But then $B$ has an element, so $B$ is not empty. So, by contraposition, if $B$ is the empty set then $B$ is a subset of each set $A$.

In order to change this, you would need to redefine "subset" or "empty set" - but then these would not refer to the concepts that we normally have for them.

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There is a logical error in your question. If $\emptyset=\{\}$, then, for any set $A$:

  • It is true that every element of $\emptyset$ is in $A$, and
  • It is true that every element of $\emptyset$ is not in $A$,

but

  • It is not true that not every element of $\emptyset$ is in A. (Watch the word order!)

This is because $(\forall x)\neg P(x)$ is not equivalent to $\neg(\forall x)P(x)$ (i.e. universal quantifier does not swap with negation). Instead, it is equivalent to $\neg(\exists x)P(x)$ (the universal quantifier changes into existential quantifier).

(Note: In your case, $P(x) := x\in\emptyset\implies x\in A$.)

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  • $\begingroup$ Thank you. But how does this address the question of "what happens if the empty set is not a subset of every set?" $\endgroup$ – étale-cohomology Jan 2 '18 at 19:46
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    $\begingroup$ Nothing happens, because it is logically impossible for an empty set to not be a subset of every set. My post was an attempt to explain where you made a mistake when you concluded that it was possible. It is not possible. $\endgroup$ – user491874 Jan 2 '18 at 20:13
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Here are some "unpleasant" consequences of your proposals.

Consider te "hypothesis":

the empty set is not a subset of any set.

This is: $\forall x \ \lnot ( \emptyset \subseteq x)$.

Applying the def of subset:

$\forall x \ \lnot \ \forall z \ (z \in \emptyset \to z \in x)$, i.e.

$\forall x \ \exists z \ \lnot (z \in \emptyset \to z \in x)$, i.e.

$\forall x \ \exists z \ (z \in \emptyset \land z \notin x)$.

This is true for every set $x$, and thus also of $\emptyset$:

$\exists z \ (z \in \emptyset \land z \notin \emptyset)$.


Consider now:

the empty set is not a subset of some sets.

This is: $\exists x \ \lnot ( \emptyset \subseteq x)$.

Applying the def of subset:

$\exists x \ \lnot \ \forall z \ (z \in \emptyset \to z \in x)$, i.e.

$\exists x \ \exists z \ (z \in \emptyset \land z \notin x)$.

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No, it is not just a convention. And the statement "there exists a set $A$ with the property that not every element of $\{\}$ is an element of $A$" is never true.

Let $P$ be the property of being an element of $A$. Now every element of the empty set has property $P$ and not of property $P$, vacuously. The first part implies that the empty set is a subset of $A$ by the definition of being a subset.

But that every element of the empty set does not have property $P$ does not imply that the empty set is not a subset of $A$. The statement that the empty set is a subset of $A$ means that every element of the empty set (there are none) is an element of $A$. But that the empty set is not a subset of $A$ means that not all elements of it satisfy $P$. But if not all elements satisfy $A$, there must be an element that satisfies not $P$. In particular, there must be an element in the empty set- which is of course not true. All elements of the empty set satisfy not $P$, but that is not enough because there are no elements.

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Those other statements that you ask about are not vacuously true. A statement is vacuously true if it has the form $\forall x(P(x)\implies Q(x))$ and in fact there does not exist $x$ such that $P(x)$ is true. The statements you ask about are simply not of this form. ("not every element..." is not of the form $\forall x\dots$, it is of the form $\neg\forall x\dots$..)

Since they're simply not vacuously true, no we can't "say" they're vacuously true unless we want to see the universe explode...

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  • $\begingroup$ Thanks for the clarification I'll delete my comments. $\endgroup$ – DanielV Jan 2 '18 at 18:34
  • $\begingroup$ I've never heard the expression "vacuously false" but think it would apply. $\lnot (\forall x: Q(x))$ and its equivalent $\exists x:Q(x)$ are both definitely false. ... Oh, I guess they are not vacuously false because they are directly and demonstrably false... or maybe all false statements are vacuously false... Well, both those statements are clearly false and definitely not true if there are no $x$.... $\endgroup$ – fleablood Jan 2 '18 at 18:46
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" if A is an arbitrary set, then not every element of {} is an element of A. Couldn't this statement be vacuously true because {} has no elements?"

No. That would make it vacuously[$*$] false.

If there are no elements anything you say about "all elements" is true. And if there are no elements anything you say about "not all elements" is false.

"Not all elements" is an exception. And for there to be an exception... there must be something to begin with.

"Not all element of V are Q" means that there must be at least one element of V that is not Q. That can not happen if there are no elements of V at all.

That makes it vacuously false; not true.


Your confusion arises in that usually the statements "All elements are Q" and "No elements are Q" are incompatible. And furthermore usually "No elements are Q" must imply "Some elements are not Q".

So it's definitely true that no elements of {} are in A. So usually that would mean is isn't possible that all elements of {} are in A..

And it also usually be true that some elements of {} are not in A.

Either one of those conclusions would mean $\{\}\not \subset A$.

But the incompatibility and implication of those statements are only true when there exists at least one element! If there are no elements then those assumptions are simply false. Not just simply false-- spectacularly false.

If there are zero elements then the statements "All elements are Q" and "No elements are Q" are not incompatible. In fact, they are completely EQUIVALENT. "No elements" is "All elements". That sounds like a semantic joke, but it's logically sound.

And "No elements are Q" does not imply "some elements are not Q". It implies "among the elements some are not Q". And if there are no elements "among the elements" means exactly "among none of the elements" (because "all elements" = "no elements").


So....

it IS true that:

a) All of the elements of {} are not in A, and

b) among all the elements of {} some would not be in A.

Those two statements are perfectly true.

But NEITHER of those statements mean $\{\} \not \subset A$.

Unless {} had at least one element.

If {} had any elements and those statements were true then {} wouldn't be a subset of A.

(But, then again, if {} had any elements we wouldn't know offhand if those statements were true or not.)

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[$*$] Maybe "ontologically false" is a more accurate term than "vacuously false".

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All the interesting discussion herewith reminds the centuries old discussion on whether $0$ is a (natural) number. I am not a specialist on either of the two subjects, but it seems to me that their discussion follows a common path. So allow me to make such a hystorical recall.

The venerable greek Scholars (in the majority) did not accept $0$ to be a number.
Later the positional notation made of it a "useful" numerical symbol, at the least, but its nature was still much debated.

Even Peano, at first, constructed his axioms starting with $1$, but later it became generally accepted that the first of the axioms be $0$ is a natural number, why ?
Wikipedia just affirms
"Peano's original formulation of the axioms used 1 instead of 0 as the "first" natural number.[7] This choice is arbitrary, as axiom 1 does not endow the constant 0 with any additional properties. However, because 0 is the additive identity in arithmetic, most modern formulations of the Peano axioms start from 0."
Is that sufficient for us, in present times, to close the discussion ?

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  • $\begingroup$ As late as the 1500s there seems to have been some controversy left about whether one is a number. $\endgroup$ – Henning Makholm Jan 2 '18 at 19:35

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