Why is a one element set not equal to the element?

Question

In treatments of set theory, it is assumed that a one element set is something different from the set itself. However, this doesn't seem obvious to me. For instance, an army containing only one soldier is the same thing as the soldier. Why can't this analogy be applied to sets?

An army constaining only one soldier John Doe is not the same as the soldier John Doe. For example, the soldier John Doe may be married whereas an army is not a person that could be married. — Hagen von Eitzen, Jul 05 '21 at 15:32
Are you really asking this from a mathematical point of view? There is a position in philosophy called Mereology which deals exactly with this problem. You may want to repost your question on Philosophy SE since there may be folks who are more into this theory and can give answers from another perspective. — rexkogitans, Jul 06 '21 at 05:59
The accepted answer points out that ${x}=x$ can never hold because of the Axiom of Regularity. The Axiom of Regularity is unimportrant for most applications of set theory to ordinary mathematics; we could drop it with no great loss. In that case, there could be sume instances where ${x}=x$. But your question seems to be asking, why isn't ${x}$ always equal to $x$? Well, what if $x$ is a $5$-element set? How can a $5$-element set be equal to a one-element set? Do you think no set should have more than one element? — bof, Jul 24 '21 at 23:47

score 16 · Answer 1 · answered Jul 05 '21 at 15:29

16

A bag containing an apple isn't the same thing as an apple.

answered Jul 05 '21 at 15:29

Hans Lundmark

53,395

That's only because there's a bag, the bag being a separate entity makes the two scenarios different. But when considering sets, we don't introduce any new entity to the elements. – Toba Jul 05 '21 at 15:33
11

A set can be thought of as a "container", though not a physical container, only a conceptual one. – Robert Israel Jul 05 '21 at 15:35
1

On what basis would we assume that we can think of sets as containers? The formal definition of sets doesn't make any reference to sets being similar to conceptual containers. – Toba Jul 05 '21 at 15:38
10

@Toba The formal definition of sets also doesn't mention anything about armies. To get better at mathematics, one must make formal reasoning from the axioms, and then try and develop personal intuitions and check them against the formal reasoning. You have a faulty intuition, leading you to conclude intuitively a statement which is formally proven false. Hence you must correct your intuition, so that it will be more useful in predicting what results the formal reasoning will give. It is not bad to start with a faulty intuition, but it is important to learn how to correct your intuition. – Jul 05 '21 at 15:43
3

I feel like Toba is really striking right at the problem with this answer. The whole point of a set, mathematically, is that it has no properties other than it elements. In particular, there is no bag. The set isn't a bag plus its contents, it's just the contents, and nothing more. That said, I think Toba is making a mistake by asking this question on Math SE, because this is a philosophical issue, and most mathematicians are not interested in philosophical issues, or even regard them as an annoying distraction. Toba should take the question to a philosopher of mathematics instead. – MJD Jul 05 '21 at 15:56
Exactly. If a set has no other properties other than that of its elements, how then is it different from the element it contains in a case where it contains one element? – Toba Jul 05 '21 at 16:06
3

@MJD If you want to be philosophical about it (which I often find mathematicians, particular mathematical logicians, succeed in better than philosophers of mathematics) then the "whole point of a set" is that it is something to which the memebership relation can meaningfully be applied; no more, no less; that is, a set is something about which it can asserted whether "something is a member of it". This is not at all the same as "having no properties other than its elements" and indeed I would say carries the 'bag' idea along with it. – Judy N. Jul 06 '21 at 00:32
2

But of course as has been mentioned in answers: whilst we may allow some set to be equal to the one-element set containing itself, depending on what set theory we use, we can also give the standard reply to any undergrad who makes this complaint which is to consider $\emptyset$ and ${\emptyset}$. How do I convince an undergrad that $\emptyset\neq{\emptyset}$? I ask them how many elements are in the first thing, and how many in the second – Judy N. Jul 06 '21 at 00:34
2

(I think just clarifying my first comment, as I'd be running it over and over in my head if I didn't): I'm saying I would rewrite your "has no properties other than its elements" to something like "has no properties other than the property of membership (of it)", and which I say are (philsophically) very much not the same: that is, considering a set $Y$ and an element $x$, I would say that "$x$", respectively, "$x\in Y$", are very different things – Judy N. Jul 06 '21 at 00:49
1

The notion that "a set can be a member of itself" leads to a logical contradiction (as Russell discovered). If a set can be a member of itself, we should be able to talk about the "set of all sets which are members of themselves" and also "the set of all sets which are not members of themselves". But the second idea is self-contradictory. If $X$ is the set of all sets which are not members of themselves, is $X$ a member of $X$? If it is, then by definition it isn't. If it isn't, then by definition it is. – alephzero Jul 06 '21 at 01:59
There is no problem with a set being an element of itself, and consistent models of non-well-founded set theory can and do exist. The problem that leads to the Russell paradox is much deeper, and comes from the unrestricted comprehension axiom. – MJD Jul 12 '21 at 19:40

score 12 · Answer 2 · answered Jul 05 '21 at 15:29

12

Mathematics doesn't work on analogies. It works on pure formal definitions. Two sets $A,B$ are equal if and only if every element of $A$ is an element of $B$, and every element of $B$ is an element of $A.$ The sets $A = \emptyset$ and $B = \{\emptyset\}$ are not equal, since $\emptyset\in B$ but $\emptyset\not\in A.$ Sets are purely formal rigorous objects. You cannot make random analogies to reason with them.

answered Jul 05 '21 at 15:29

1

So is the fact that a one element set is not equal to the element a formal axiom? Or it can be proved? – Toba Jul 05 '21 at 15:45
2

@Toba It is not an axiom. It can easily be proved from the axiom defining equality of sets, as I illustrated in my post. In general, if you take a set $A$ and a set $B = {A},$ then $A\in B,$ but $A\not\in A$ (this is the axiom of regularity). Hence $A \neq B.$ I chose a more concrete case, to avoid explaining the trickier point of why $A\not\in A$ for every set $A.$ – Jul 05 '21 at 15:47
2

@Toba The empty set is a set. This is easily provable from the axioms. For instance, the axiom of infinity tells us that there is a set $N$ (it stipulates other things about this set, but we won't need them), so therefore there exists a set $\emptyset = {a \in N \mid a \in a}.$ You can then prove, as I mentioned above, that there is no $a$ so that $a \in \emptyset,$ therefore there is a set containing nothing. – Jul 05 '21 at 15:48
I think I understand this special case now. It is due to the fact that the element of B is a set and a set is not an element of itself. But how do we extend the proof to cases where the element is not a set? – Toba Jul 05 '21 at 15:58
@Toba In set theory, everything is a set. – Jul 05 '21 at 16:00
1

I don't agree with that statement. Taking the set of all integers for instance, 5 is an element of that set. 5 is not a set. Though we can of course define a set where 5 is the only element, that still doesn't change the fact that 5 is not a set. – Toba Jul 05 '21 at 16:04
3

@Toba 5 is a set in set theory. In typical treatments, the natural number 5 is the set ${0, 1, 2, 3, 4}.$ In set theory, everything has to be a set (how else do you define anything?). We just rarely think of 5 as being a set, since it's not very useful (and oftentimes counterproductive). – Jul 05 '21 at 16:05
So are you saying that even in cases where the element is not a number; for instance, the set of all dogs on Earth, each element is still a set? Even though we haven't defined them as such? – Toba Jul 05 '21 at 16:11
2

@Toba What is the set of all dogs on earth? You can try intuitively reasoning about such an object, but until you have a formal definition of 'dog' and 'earth,' it is complete bogus to claim the rules of set theory apply to such an object. – Jul 05 '21 at 16:12
Is the fact that all elements of a set are themselves sets an axiom in set theory? – Toba Jul 05 '21 at 16:22
@Toba This is implicit in the formalization of set theory. If you look at a formal treatment of set theory, you will not find "for all sets A, B, C, ...," you will find the axioms stated universally as "For all x, y, z, ...". In any formula you write in set theory, the objects are sets. The axioms of sets apply to everything you talk about. – Jul 05 '21 at 16:27
According to the answer given in the link below, not all elements of sets are themselves sets: https://math.stackexchange.com/a/1709439/828305 – Toba Jul 05 '21 at 16:41
1

@Toba That comment is proposing you go outside the framework of set theory. If you leave set theory, of course not everything is a set. If you are in set theory, everything is a set. – Jul 05 '21 at 16:59
I think I understand now, set theory assumes everything that exists is a set. – Toba Jul 05 '21 at 20:19
1

@Toba as a further point, you have misunderstood what that answer is saying. It is making the distinction between a model and the 'thing' you model; it makes no reference to whether an element of a set is a set. – masiewpao Jul 06 '21 at 00:50

Somos · Accepted Answer · 2021-07-09T10:56:10.103

You are asking why $\,x=\{x\}\,$ is false in set theory. The usual answer is that $\,x\in\{x\}\,$ by definition of singleton sets and thus, if $\,x=\{x\}\,$ were true, then this would imply that $\,x\in x\,$ which is usually forbidden by the axiom of regularity in Zermelo-Fraenkel set theory. However, the Wikipedia article also states

Virtually all results in the branches of mathematics based on set theory hold even in the absence of regularity

and thus it is a matter of convention and of convenience. The article further states

In addition to omitting the axiom of regularity, non-standard set theories have indeed postulated the existence of sets that are elements of themselves.

and thus it is depends on which particular version of set theory you are using. Consult the Wikipedia article Non-well-founded set theory for some of these details.

This is further complicated by the existence of urelements. The axiom of regularity article further states

Urelements are objects that are not sets, but which can be elements of sets. In ZF set theory, there are no urelements, but in some other set theories such as ZFA, there are.

and again, it depends on which particular version of set theory you are using. The key idea is that set theory is a theory with many different variations and it sometimes matters which one you decide to use.

Your question about

Why can't this analogy be applied to sets?

is that it could -- provided that you use some particular consistent version of set theory which agrees with your intuition.

Tanner Swett · Answer 4 · 2021-07-24T23:30:20.890

Suppose that $S$ is the set $\{1, 2, 3\}$. Is there a set which has $S$ as an element, but contains no other elements?

Maybe there is, maybe there isn't. Let's consider both possibilities.

If the answer is yes

In this case, yes, there is a set which has $S$ as an element, but has no other elements. Let's call that set $\{S\}$.

Is it possible that $\{S\} = S$? Well, we know that $\{S\}$ has only one element (namely, $S$). On the other hand, $S$ has multiple elements. So $\{S\}$ cannot possibly be the same thing as $S$.

If the answer is no

In this case, no, there is no set which contains $S$ as an element but has no other elements.

If you want to define sets this way, then you have every right to do that! But now working with sets is going to become very frustrating.

Why would it be frustrating? Well, suppose you want to say, "Let $T$ be the set of all sets which have such-and-such property." What happens if if $S$ has that property and no other set has that property? Bad things happen! Let me explain.

Since $T$ is the set of all sets which have such-and-such property, and $S$ is the only set with that property, we can conclude that the set $T$ has the set $S$ as an element, but $T$ has no other elements. But wait a minute—we just said, a few paragraphs ago, that that's impossible! So, apparently, our definition of $T$ is invalid. That's awfully confusing.

We mathematicians don't want to have to face those kinds of frustrations, so when we define sets, we choose to define them in such a way that $\{x\}$ does not have to equal $x$. Indeed, most mathematicians choose to define sets in such a way that $\{x\}$ is never equal to $x$. That seems to make things easiest.

Chris Bouchard · Answer 5 · 2021-07-06T02:19:18.177

I think you're getting tripped up because you're trying to use concepts from “set theory” ¹ directly on real world things. Mathematics does not operate in the real world — it operates in the world of axioms and statements and proofs.

Digression into Formal Logic

A theory is nothing more or less than:

a selection of symbols: operators, function symbols, variables, etc., and
axioms: statements made using those operators, function symbols, and variables, combined with symbols defined by the logic being used, often First-Order Logic, a.k.a Predicate Logic (as is the case for set theory).

These are all purely theoretical and symbolic objects. They have no meaning (at least no inherent meaning). So, for set theory, we have the operator $\in$, the constant symbol $\emptyset$, and an infinite set² of variables.³

The axioms are statements, like the axiom of extensionality in set theory: $$\forall x \forall y [ \forall z ( z \in x \iff z \in y) \implies x = y].$$ The symbols $\forall$, $\iff$, $\implies$, and $=$, are all defined by First-Order Logic. We can construct other statements, purely symbolically, from the grammar of the operators, function symbols, and variables. These statements also have no inherent meaning.

From a statement, we can attempt to prove or disprove the statement, by constructing various proofs using the logic system combined with the axioms. The proof itself is (again) purely symbolic. If we can prove the statement, we say it is “true”, and if we can disprove it we say it is “false.”

I mention repeatedly that these objects are all purely symbolic because it's important to understand that the theory by itself is useless, at least for the purpose of describing sets. There isn't even anything called a set in the universe of set theory — only operators and statements and axioms and truth values. It's only called “set theory” because we have in mind a particular use for it.

Back to Reality

So if theories are purely symbolic and meaningless, what good is set theory, or indeed formal mathematics? The answer is that we pick other systems, called models, and map the theory onto the model. We provide a mapping from symbolic objects to objects in the model, and show that the axioms are “true” ⁴ for the model. We say that the model is an interpretation of the theory, and then we know for any formal statement in our theory, there's a corresponding statement about the model with the same truth value. (I'm hand-waving here.)

So why is $x \neq \{x\}$?

Because, at the end of the day, if we want to apply the theory of set theory to our model of sets, we need to choose a model of sets that interprets the theory of set theory — and in set theory, the statement $\exists x \exists y (x = y \land x \in y)$ can be disproven, thus is false, and thus the corresponding statement must be false in our model. So we know your model, where an army of a single soldier is equal to the soldier themselves is not a model of set theory.

So why did the creators of set theory choose for their theory to work this way? As I've said, the theory has no inherent meaning, so its creators had a countably infinite universe of theories to choose from — so this was a real choice. I think probably because they found their theory to be a good compromise of:

intuitive: having models that most mathematicians can understand,
useful: having the sorts of models that mathematicians would need in their work,
powerful: having axioms that can capture the truth value of many different sorts of statements, and
tractable: having axioms that actually allow most useful statements to be proven.⁵

As others have said, it's possible to define other theories that capture the property you've mentioned, but presumably they have trade-offs in these areas. We've chosen to build a lot of our mathematics on top of the modern formulation of set theory, so it's been Good Enough™.

¹ I put the phrase in scare quotes because there have been multiple formulations of theories of sets over the ages. The most popular today is Zermelo–Fraenkel set theory, and it's the one most people mean when they say “set theory.”

² This is of course slightly begging the question, because we're defining a theory of sets using sets, but there's a distinction. The set of variables is not part of the universe of sets being described by the theory. It's purely part of the formalization of the theory.

³ Note that the brace operator $\{\cdot\}$, a.k.a. set-builder notation, is not part of this theory. It's defined purely in terms of $\in$ and $=$.

⁴ More scare quotes because, as you may see, this is a recursive definition. We'd need to use some formal logic system to prove this correspondence. There's a reason Whitehead and Russel's Principia Mathematica only completes the proof that $1 + 1 = 2$ in the second volume.

⁵ Gödel's Incompleteness Theorem shows that any “suitably powerful” theory in first-order logic is either incomplete (has statements that cannot be proven or disproven) or inconsistent (has statements that can both be proven and disproven). So the goal of a theory is to make many useful statements provable and avoid being inconsistent. This is part of the “craft” of formal logic.

score 1 · Answer 6 · answered Jul 06 '21 at 04:20

1

Suppose $A$ is the set $\{1, 2, 3\}$, and $B$ is the set $\{A\}$. Then $B$ contains only one element, namely $A$, but $A$ contains three elements. So they can’t possibly be the same thing — they contain different numbers of elements!

answered Jul 06 '21 at 04:20

mweiss

23,647

1

Ah, I see that my answer is almost identical to Tanner Swett’s. Ah well. – mweiss Jul 06 '21 at 04:21