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Recently, I learned that for any set $A$, we have $\varnothing \subset A$. I found some explanation of why it holds.

$\varnothing\subset A$ means "for every object $x$, if $x$ belongs to the empty set, then $x$ also belongs to the set $A$". This is a vacuous truth, because the antecedent ($x$ belongs to the empty set) could never be true, so the conclusion always holds ($x$ also belongs to the set $A$). So $\varnothing \subset A$ holds.

What confused me was that, the following expression was also a vacuous truth.

For every object $x$, if $x$ belongs to the empty set, then $x$ doesn't belong to the set $A$.

According to the definition of the vacuous truth, the conclusion ($x$ doesn't belong to the set $A$) holds, so $\varnothing \not\subset A$ would be true, too. Which one is correct? Or is it just a convention to let $\varnothing\subset A$?

Rodrigo de Azevedo
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Searene
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    Maybe think of it like this: The empty set is like an empty bag. The set $A$ is like a bag with some stuff in it. It's possible to put your hand into either bag and take nothing out. Since everything you could take out of the empty bag $\varnothing$ was also in $A$, this means that $\varnothing \subset A$. This argument may need to be made rigorous, but I think the intuition is a good start. – jamesh625 Oct 04 '16 at 09:31
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    It's a consequence of the following rule: any statement of the form 'if $x \in \emptyset$, then y' is true. That rule is itself a consequence of the agreement that "if P then Q" means "P is false or Q is true." On a philosophical level, I would distinguish this agreement from being a mere convention since it's an agreement about what reasoning itself means. But on a practical level, if you get annoyed thinking about these issues in any particular context, it's often better to shrug your shoulders, call it a convention, and move on! – hunter Oct 04 '16 at 11:53
  • (the previous comment doesn't answer your specific question, but only the question in the title, which is why I haven't made it an answer.) – hunter Oct 04 '16 at 11:53
  • When you say "I have no money" think of the no money as the currency that everyone has. – polfosol Oct 05 '16 at 06:14
  • And if I borrow you no money, what would be the total amount of money you have afterwards? – polfosol Oct 05 '16 at 06:25
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    It's the way math works... It is not a convention, but a theorem: we have an axiom asserting that there is a unique set $y$ such that : $\forall x \lnot (x \in y)$ and we call $y$ "the emptyset". Then we assume the def of set-inclusion : $A \subset B \leftrightarrow \forall x (x \in A \to x \in B)$ and we derive, by logical rules, that $\emptyset \subset A$, for any $A$. – Mauro ALLEGRANZA Oct 05 '16 at 10:15
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    It is true that for every $x$ and every set $A$, if $x$ belongs to the empty set, then $x$ both belongs to $A$ and $x$ does not belong to $A$ – Henry Oct 06 '16 at 00:32
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    “For every object $x$, if $x$ belongs to the empty set, then $x$ doesn’t belong to the set $A$.” This statement is vacuously true, indeed. However, it does not imply that $\varnothing\not\subset A$. For the latter to be true, there should actually exist some $x$ such that $x\in\varnothing$ but $x\notin A$. – triple_sec Oct 06 '16 at 22:56
  • It is either an axiom or a derivable theorem, not a 'convention'. In this case it is a derivable theorem. – user207421 Oct 07 '16 at 00:24

7 Answers7

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There’s no conflict: you’ve misinterpreted the second highlighted statement. What it actually says is that $\varnothing$ and $A$ have no element in common, i.e., that $\varnothing\cap A=\varnothing$. This is not the same as saying that $\varnothing$ is not a subset of $A$, so it does not conflict with the fact that $\varnothing\subseteq A$.

To expand on that a little, the statement $B\nsubseteq A$ does not say that if $x\in B$, then $x\notin A$; it says that there is at least one $x\in B$ that is not in $A$. This is certainly not true if $B=\varnothing$.

Brian M. Scott
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  • I wouldn't say he misinterpreted the second statement; his interpretation is correct. But that is not the definition of $\nsubseteq$ – BlueRaja - Danny Pflughoeft Oct 04 '16 at 17:32
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    @BlueRaja-DannyPflughoeft: Which is to say that he misinterpreted it: it does not mean what he understood it to mean. – Brian M. Scott Oct 04 '16 at 17:47
  • Now I have doubts as to whether it is correct to use vacuous truth to interpret anything in the first place. E.g. consider the following expression: For every object of x, if x belongs to the empty set, then the empty set is not a subset of A. This is also a vacuous truth, so the conclusion(the empty set is not a subset of A) should be true. In fact, because it's a vacuous truth, the conclusion should be true no matter what it's expressing. Another example, If 1=2, then 5=6. Should I say that 5=6 is true? – Searene Oct 05 '16 at 06:38
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    @Searene You can derive anything from a contradiction. This is basic logic, and nothing wrong with it. – orlp Oct 05 '16 at 07:17
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    No you shouldn't say that 5=6 is true, but you can say that If 1=2, then 5=6 is true. The statement that x belongs to A is not necessarily true; what is true is precisely: for every object x, if x belongs to the empty set, then x also belongs to the set A, and that is the definition of the empty set being a subset of A. – Ari Brodsky Oct 05 '16 at 08:13
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    @Searene The definition of inclusion is If x belongs to A, then x belongs to B. For the empty set, it is indeed a vacuous truth, which does prove that the empty set is a subset of any set. But the proof is the whole implication, it is not the second proposition (which is irrelevant because the first proposition is false). – T. Verron Oct 05 '16 at 09:39
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    @Searene For every object of x, if x belongs to the empty set, then my pet lion is Barack Obama's favourite crescent wrench is also true. (given an appropriate formalism of "my pet lion is Barack Obama's favourite crescent wrench") – Stack Exchange Broke The Law Oct 05 '16 at 09:50
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    @Searene You are making a fundamental error. You cannot derive that the conclusion is true!!! If you have a statement: $\forall x, x \in \emptyset \rightarrow \emptyset \not\subset A$ is true because $x \in \emptyset$ is false. This tells you nothing regarding $\emptyset \not\subset A$. In an implication $P \rightarrow Q$ if $P$ is false than the whole implication is true, but this does not say anything about the truth value of just $Q$. – Bakuriu Oct 05 '16 at 13:02
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    @Searene: The underlying problem is that you’ve misunderstood what vacuously true means. A statement of the form $\forall x,(P(x)\to Q(x))$ is vacuously true if $\forall x,P(x)$ is false. In words, the first says that every $x$ with property $P$ has property $Q$, and the second says that there is no $x$ with property $P$. Since there is no $x$ with property $P$, the implication $P(x)\to Q(x)$ is true for every $x$, irrespective of whether any $x$ has property $Q$: $Q$ never actually comes into play at all. Your statement $\forall x,(x\in\varnothing\to\varnothing\nsubseteq A)$ is ... – Brian M. Scott Oct 05 '16 at 17:23
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    ... vacuously true, but that doesn’t tell you anything about whether or not $\varnothing$ is a subset of $A$. The truth of $\forall x,(x\in\varnothing\to x\in A)$, on the other hand, means by the definition of $\subseteq$ that $\varnothing\subseteq A$. – Brian M. Scott Oct 05 '16 at 17:25
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    Thank you, everyone. I've read all the comments above, and I think I've understood what you mean. Vacuous truth is also a truth, although the antecedent could never be true. So most of them don't really mean anything. What makes $\forall x(x\in\varnothing\to\varnothing\in A)$ special is that. it happens to be the definition of $\subset$, so $\forall A$, $\varnothing\subset A$ holds – Searene Oct 05 '16 at 23:40
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    @Searene Don't you mean $\forall x ( x \in \varnothing \to x \in A)$? – Kyle Strand Oct 06 '16 at 18:40
  • @Brian: You mean that $\forall x(P(x)\to Q(x))$ is vacuously true if $\exists x,P(X)$ is false, right? (Or, equivalently, if $\forall x,\neg P(x)$ is true). – hmakholm left over Monica Jan 16 '17 at 13:56
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From Halmos's Naive Set Theory:

enter image description here

A transcription:

The empty set is a subset of every set, or, in other words, $\emptyset \subset A$ for every $A$. To establish this, we might argue as follows. It is to be proved that every element in $\emptyset$ belongs to $A$; since there are no elements in $\emptyset$, the condition is automatically fulfilled. The reasoning is correct but perhaps unsatisfying. Since it is a typical example of a frequent phenomenon, a condition holding in the "vacuous" sense, a word of advice to the inexperienced reader might be in order. To prove that something is true about the empty set, prove that it cannot be false. How, for instance, could it be false that $\emptyset \subset A$? It could be false only if $\emptyset$ had an element that did not belong to $A$. Since $\emptyset$ has no elements at all, this is absurd. Conclusion: $\emptyset \subset A$ is not false, and therefore $\emptyset \subset A$ for every $A$.

Rodrigo de Azevedo
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    This paragraph shines with the rare virtue of sympathy with (and generosity of spirit toward) the uninitiated. +1. – Jason Orendorff Oct 04 '16 at 15:16
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    I love this book. – Andreas Caranti Oct 04 '16 at 15:27
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    This answer isn't screen-reader friendly, it's better to post actual text. – djechlin Oct 04 '16 at 18:32
  • @djechlin How? Google Books does not allow copying text, to the best of my knowledge. – Rodrigo de Azevedo Oct 04 '16 at 19:02
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    The old fashioned way works. –  Oct 04 '16 at 19:09
  • Advice: if the image in my answer isn't properly displayed in your device, click on the link to go to the Google Books page and search for "vacuous" inside the book. It's how I found the relevant page. – Rodrigo de Azevedo Oct 04 '16 at 19:20
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    sigh, done. .. – djechlin Oct 04 '16 at 19:21
  • @djechlin If there are copyright violation issues, it's your transcription, not mine. – Rodrigo de Azevedo Oct 04 '16 at 19:25
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    With the text as opposed to with a picture of the text? – djechlin Oct 04 '16 at 19:41
  • @djechlin Text is much easier for search engines to digest. – Rodrigo de Azevedo Oct 04 '16 at 19:46
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    Okay, I don't see what search engines have to do with copyright violation issues, but in the meantime, I'd rather see the answer be accessibility compliant, and I guess it's your legal understanding that I'm responsible for any "issues" that might come up, so okay. – djechlin Oct 04 '16 at 19:48
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    I like Halmos's second demonstration because it highlights a frequently used technique: For an implication to be false, there must be a witness to its falseness. That is, there is an element of some set that is a counterexample. Conversely, if it is impossible that there are counterexamples, then the implication is true. – Eric Towers Oct 05 '16 at 12:59
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    Stack Exchange is one of those extremely rare cases where almost all, if not all, copying of material actually does constitute fair use (in US law, at least). It seems to me that it would be trivial to establish the fact that quotes on SE are for commentary and/or educational purposes, both of which are considered fair use. – Todd Wilcox Oct 07 '16 at 03:31
  • Can't we argue the other way?

    That is, to show that something is false, we have to show that it is impossible for it to be true. Since there are no elements in the set, clearly it can satisfie no conditions, and is therefore false.

     – sloth Oct 21 '16 at 02:41
  • @sloth Translate that to predicate logic. – Rodrigo de Azevedo Oct 21 '16 at 09:22
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What confused me was that, the following expression was also a vacuous truth.

For every object of $x$, if $x$ belongs to the empty set, then $x$ doesn't belong to the set $A$.

As a complement (heh) to Brian Scott's (+1) answer, your argument shows that $\varnothing \subset A^{c}$, the complement of $A$. This statement is also (vacuously) true.

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Andrew D. Hwang
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Very subtle point:

"All x are not something" does not imply "Not all x are something".

The first may be vacuously true. The second one can not. If the x are vacuous then the second one has to be "vacuously false" as all x of nothing are any property so it is impossible for them not to be any property.

So "all elements of the empty set are not in $S$" does not imply "Not all elements of the empty set are in $S$" $\iff$ "It is not true that all elements of the empty set are in $S$" $\iff$ "There are some elements of the empty set that are not in $S$".

The first is vacuously true (and is equivalent to $\emptyset \subset S^c$ which is true) and the second set of equivalent statements are all equivalent to $\emptyset \not \subset S$ which is not true.

=====

The thing is what you say is absolutely correct for non empty sets.

More formally:

All elements $x$ in $S$ are not in $A$ $\implies$

$S \subset A^c$ $\implies$

$\color{red}{\text{There is an } x\in S \text{ where } x \not \in A}\implies$

It is not true that all $x \in S$ are also in $A$ $\implies$

$S \not \subset A$.

However the red line can only be concluded if $A$ is non-empty. If $A$ is empty the red line is simply false.

And without the red line there is simply no logic or means to jump from the line before to the line after:

All elements $x$ in $S$ are not in $A$ $\not\implies$

It is not true that all $x \in S$ are also in $A$.

That simply is not true for an empty $S$.

fleablood
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Every theory has axioms, which are some propositions held to be true without being proven from anything else, and are not provable from each other. Subsequent truths of the theory derived from the axioms are theorems.

The properly termed question is whether the empty set being a subset of every other set is axiom of set theory, or a theorem.

It depends on how "subset" is defined. If $A\subset B$ means that every element of $A$ is in $B$, it is not necessarily true that $\emptyset$ is a subset of anything, since it has no elements. In this case, $\emptyset \subset A$ can be added as an axiom. It doesn't conflict with anything, and simplifies all reasoning about subsets. Alternatively, if $A\subset B$ is defined as "$A$ has no elements that are not also in $B$", then we do not require the extra axiom for the $\emptyset$ case. If $A$ has no elements at all, it has no elements that are not in $B$.

Suppose that we use the first, positively termed definition of subset, and then adopt as an axiom not $\forall A:\emptyset \subset A$, but rather its negation: $\exists A:\emptyset \not\subset A$, or the outright proposition $ \forall A:\emptyset \not\subset A$.

This is just going to cause problems. We can "do" set theory as before, but all the theorems will be uglified by having to avoid the special cases involving the empty set. In any derivation step in which we rely on a subset relation being true, or assert one, we will have to add the verbiage of an additional statement which asserts that the variable in question doesn't denote the empty set. This proposition then has to be carried in all the remaining derivations, unless something else makes it superfluous (some unrelated assurance from elsewhere that the set in question isn't empty).

Working with this clumsy subset definition that doesn't work with the empty set very well, someone is eventually going to have an epiphany and introduce a new subset-like relation which doesn't have these ugly problems: a new $A\ \mathbf{subset*}\ B$ binary relation which reduces exactly to $A\subset B$ when neither $A$ nor $B$ are $\emptyset$, and which, simply by definition, reduces to a truth whenever $A = \emptyset$, regardless of $B$. That person will then realize that all the existing work is simpler if this $\mathbf{subset*}$ operation is used in place of $\subset$.

At the end of the day it boils down to criteria like: is the system consistent (doesn't contradict itself), is it complete (does it capture the truths we want) and also is it convenient: are the rules configured so that we do not trip over unnecessary cases and superfluous logic.

Kaz
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This is a very mundane explanation (with finite sets). A subset is made of any combination of elements from the set. Suppose a set$ S$ is made of three elements: $\{a,b,c\}$. Each element $a$, $b$ or $c$ can be, or not, in the combination. If all of them are not in the combination, they STILL form a combination of "absent elements", or $\emptyset$, a subset of $S$. They are the dual of the subset of "all elements", $\{a,b,c\}$.

In the same way that $\{a,b\}$ and $\{c\}$ are (complementary) subsets, $\emptyset$ and $\{a,b,c\}$ belong to the set of subsets.

Laurent Duval
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What confused me was that, the following expression was also a vacuous truth.

For every object of $x$, if $x$ belongs to the empty set, then  $x$ doesn't belong to the set $A$.

The contrapositive of the above conditional is:

"For every $x$, if $x$ belongs to set $A$, then $x$ doesn't belong to the empty set" which is easy to understand as the empty set has no elements at all.

NOTE- $p\rightarrow q$ has same meaning as $\lnot q\rightarrow \lnot p$

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Nitin Uniyal
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