Is "The empty set is a subset of any set" a convention?
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 $\varnothing\subset A$?
elementary-set-theory logic
Harry Richman
SeareneSearene
$\begingroup$ 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. $\endgroup$ – jamesh625 Oct 4 '16 at 9:31
$\begingroup$ 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! $\endgroup$ – hunter Oct 4 '16 at 11:53
$\begingroup$ (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.) $\endgroup$ – hunter Oct 4 '16 at 11:53
$\begingroup$ 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 $A$. $\endgroup$ – Mauro ALLEGRANZA Oct 5 '16 at 10:15
$\begingroup$ 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 $A$ $\endgroup$ – Henry Oct 6 '16 at 0:32
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. ScottBrian M. Scott
$\begingroup$ @BlueRaja-DannyPflughoeft: Which is to say that he misinterpreted it: it does not mean what he understood it to mean. $\endgroup$ – Brian M. Scott Oct 4 '16 at 17:47
$\begingroup$ @Searene You can derive anything from a contradiction. This is basic logic, and nothing wrong with it. $\endgroup$ – orlp Oct 5 '16 at 7:17
$\begingroup$ 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. $\endgroup$ – Ari Brodsky Oct 5 '16 at 8:13
$\begingroup$ @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 $Q$. $\endgroup$ – Bakuriu Oct 5 '16 at 13:02
$\begingroup$ ... 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 $\varnothing\subseteq A$. $\endgroup$ – Brian M. Scott Oct 5 '16 at 17:25
From Halmos's Naive Set Theory:
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 AzevedoRodrigo de Azevedo
$\begingroup$ This paragraph shines with the rare virtue of sympathy with (and generosity of spirit toward) the uninitiated. +1. $\endgroup$ – Jason Orendorff Oct 4 '16 at 15:16
$\begingroup$ I love this book. $\endgroup$ – Andreas Caranti Oct 4 '16 at 15:27
$\begingroup$ With the text as opposed to with a picture of the text? $\endgroup$ – djechlin Oct 4 '16 at 19:41
$\begingroup$ 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. $\endgroup$ – djechlin Oct 4 '16 at 19:48
$\begingroup$ 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. $\endgroup$ – Eric Towers Oct 5 '16 at 12:59
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.
Andrew D. HwangAndrew D. Hwang
$\begingroup$ So the empty set, being a subset of both the set $A$ and its complement, is a subset of the intersection of the set $A$ with its complement, which is the empty set. $\endgroup$ – Carsten S Oct 5 '16 at 9:05
$\begingroup$ I believe this is the key point. $\endgroup$ – Kyle Strand Oct 5 '16 at 21:31
$\begingroup$ @CarstenS: Be careful that in ZF set theory the complement of a set is never a set! $\endgroup$ – user21820 Oct 7 '16 at 14:17
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 $S$.
fleabloodfleablood
$\begingroup$ You confused $A$ and $S$ at the end there. $\endgroup$ – dejongbrent Oct 6 '16 at 21:50
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 DuvalLaurent Duval
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.
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 $\lnot q\rightarrow \lnot p$
Nitin UniyalNitin Uniyal
I'm also learning about sets, but the way that I've begun to see things like this is as follows:
I think about what a nice system needs, and a few things spring to mind. I need an addition, the idea of pushing two things together. How about a concept of negatives? That could be useful. I could add these negatives to some of these positives, and eventually I'm going to come across a situation where my negatives meets my positives and what do I have? That's the null set.
Think about 'zero' logically, this concept of nullness, but more than just the number zero, but its function, and realize why it fits soundly within our understanding of mathematics. Say we were to not have this zero, the what would happen? This concept of zero is such an obvious idea in numbers that when we get zero in more bizarre systems like sets, the concept of zero along this feels odd, because 0 is just a number after all.
But in sets your '0' still retains the set construct, but it represents the analogous function that 0 holds in numbers.
SthomasSthomas
$\begingroup$ (1) I don't understand this. Do you have a concept of a negative set, that you could "add" to a non-empty set to yield the null set? If so, please explain. (2) How does this relate to the question? It seems as if you are saying that $A⊆B$ in sets corresponds to $A≤B$ in numbers, and since $0$ is $≤$ all (positive) numbers, then it is analogous that $\varnothing⊆A$ for all sets $A$. But $A⊆B$ doesn't correspond to $A≤B$; it's more like $A\mid B$ (i.e., $A$ divides $B$). And since $0$ isn't a factor of any number, you have just augmented the OP's confusion, and not alleviated it. $\endgroup$ – Scott Oct 5 '16 at 15:34
$\begingroup$ Welcome to the site; unfortunately, this is not an answer to the question asked. Answers on StackExchange sites cannot be merely commentary or discussion; that's what comments and the chat rooms are for (though unfortunately you do not have sufficient rep to write comments or participate in chat yet). Please consider taking the tour. $\endgroup$ – Kyle Strand Oct 5 '16 at 21:31
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