A question about the definition of 'set'

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A set is defined as: "A set is a finite or infinite collection of objects in which order has no significance, and multiplicity is generally also ignored (unlike a list or multiset)." (http://mathworld.wolfram.com/Set.html" )

This is my question: can it be a collection of real objects or does it have to be a collection of abstract or conceptual objects only?
.
 
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murshid_islam said:
A set is defined as: "A set is a finite or infinite collection of objects in which order has no significance, and multiplicity is generally also ignored (unlike a list or multiset)." (http://mathworld.wolfram.com/Set.html" )

This is my question: can it be a collection of real objects or does it have to be a collection of abstract or conceptual objects only?
.

I don't know why Wolfram Mathworld put a restriction on their definition re order. Sets can be partially or totally ordered. There's no reason why there can't be sets of real objects as long as the number of elements is definable. My Harper-Collins Dictionary of Mathematics (Borowski and Borwein eds) gives the following definition: "...a collection of distinct members, possibly infinite, that is treated as an entity ...with its identity dependent only on its members."
 
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SW VandeCarr said:
Sets can be partially or totally ordered.
A fact which has no significance for the set.

If we consider both a set and an ordering on it, we call it an "ordered set". Compare with the fact any nonempty set can be given a group law.
 
A mathematical set is an abstract thing, and as a mathematical term it never refers to physical things.
 
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There isn't really a definition of a set. It's an abstract thing (quoting Jarle), on which most of mathematics is built. Just think of it as "a bunch of stuff" in the most general sense. The 'objects' of a set (or 'elements' or 'stuff it's made up of') can be anything.

(Well, to some degree. You can't have a 'set of all sets'. But that's another story.)

@SW VandeCarr: order of the elements has no significance in the sense that the sets {a,b} and {b,a} are equal. Introducing an order on a set (such as saying that a<b) gives the set structure; then it isn't a bare set anymore but a set with structure.
 
A is a set iff there exists a z such that z is a member of A or A is the empty set

A set is anything that has elements (objects of discourse) or is empty.

This is the simplest definition of a set using the atomic statements of membership and the empty set.

Mathworld's definition presupposes axioms left unstated.

The term "z" stands for anything at all: meaning cats, dogs, thoughts, words, sentences, concepts, ghost particles, numbers, other sets, etc. in any state of mixture.

The axiom of pairing makes clear that the set {x} = {x,x}, so it doesn't matter if two or more of the same object appear in the set (Mathworld's definition leaves out the nitty gritty details of the axioms of set theory); nor is there any implied order {a,b}={b,a}.
 
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xxxx0xxxx said:
A is a set iff there exists a z such that z is a member of A or A is the empty set

This is a statement which makes no sense outside the context of set theory which has as ontological assumption that the universe of discourse is sets only, so it's circular as a "definition"(it's not a definition) of a set, and a tautology as a statement in set theory.

Sets doesn't have a definition, and they don't need a definition simply because they are what we axiomatize in set theory.
 
Jarle said:
This is a statement which makes no sense outside the context of set theory which has as ontological assumption that the universe of discourse is sets only, so it's circular as a "definition"(it's not a definition) of a set, and a tautology as a statement in set theory.

Sets doesn't have a definition, and they don't need a definition simply because they are what we axiomatize in set theory.

Actually, the definition defines A only, and leaves z undefined, i.e. it may be a set, but it need not be a set. (See for instance, Suppes, 1960).
 
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xxxx0xxxx said:
Actually, the definition defines A only, and leaves z undefined, i.e. it may be a set, but it need not be a set. (See for instance, Suppes, 1960).

A set is an entity which has no definitions but only axioms which it satisfies. Your definition is not really a definition and will not be accepted as such in mathematics.
I see that Suppes is a philospher, which kind of explains it, because mathematicians don't give a definition of a set.

That said, in ZF set theory, elements of a set are always sets. In fact, all is sets. There are other set theories out there with elements that are not sets (but so-called urelements), but these are not standard.
 
  • #10
Well...

We aren't even at ZF when we're talking about what a set is.

Cantor's intuitive set theory predates ZF and he defines set as:

‘any collection into a whole M of definite and separate objects m of our intuition or our thought. These objects are called the “elements” of M’.
 
  • #11
That is not a mathematical definition and hence not of mathematical interest. It may make as much or little sense to anyone, it doesn't matter when it comes to the mathematical part of set theory. In set theory it does not make sense to define a set, as any definition would necessarily assume the axioms for sets.
 
  • #12
Well, the point is, Cantor was doing naive set theory. And it has been shown that naive set theory was wrong (ie consistent). Thus any definition by Cantor that defines a set is also wrong.
99% of the mathematicians today take ZF as their set theory, and ZF doesn't give a definition of a set. You might give all kinds of definitions of sets, but the point is that mathematicians today won't really agree with you...
 
  • #13
The inconsistency of naive set theory does not invalidate all definitions in naive set theory.

Only those definitions directly dependent on a contradictory statement in naive set theory are invalid.

That said, let us see what makes the definition:

\mbox{A is a set} \Leftrightarrow \exists z ( z \in A \vee A = \emptyset)

a valid definition.

1) It is Eliminable because there is a clear recipe to eliminate the left hand statement "A is a set" from any formula by replacing it with the statement on right hand side.

2) It is non-creative, because no new result is derivable from this definition that cannot be derived using the primitive statements of the left hand side.

As such, it is a perfectly acceptable definition that says what it says, and no more.
 
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  • #14
I'm just saying that no single mathematicians accepts this definition and works with it. Mathematics nowadays comes from ZF mathematics and uses axioms of sets. You can invent all kind of pretty definitions, but I know no mathematician that uses them.

The problem with your definition is that you give no information about what z is and what it could be. At least, in ZF, you know that all is sets. But with your definition, you have mysterious objects z that I have no idea about what they should be.

Again, many philosophers invent all kind of weird definitions and things, but that doesn't mean that mathematicians actually use those things. More than 99% of mathematics happens in ZF, whether you like that or not.
 
  • #15
You're right, z is undefined, but it does not need definition at this stage.

The definition of set only states that if there is a defined z, then it may be used to define a set.

The richness of ZF is built up from this definition as the axioms introduce what z can be.
 
  • #16
xxxx0xxxx said:
The richness of ZF is built up from this definition as the axioms introduce what z can be.

This is not true. ZF does not use this definition. The concept of set and of membership are undefined objects in ZF.
 
  • #17
Alright, a simple example, without ZF we can define the natural numbers as sets:

0=the empty set
1={0}
2={0,1}={0,{0}}
3={0,1,2}={0,{0},{0,{0}}}
4={0,1,2,3} etc.
.
.

This construction does not require ZF, but it does require the definition of set.
 
  • #18
I really wouldn't know why your construction would require a definition for "set". I can easily give some axioms that allow me to do this, and which leave set undefined. In fact, this is what mathematicians do!
 
  • #19
Well,

We know 0 is a set, but what is 1? Is it a set?
 
  • #20
xxxx0xxxx said:
Well,

We know 0 is a set, but what is 1? Is it a set?

1={0}. This a set by the pairing axiom. Indeed, the pairing axiom states that when given sets y and z, we have a set {y,z}. (this is informally what it says, formally it's something quite different). Thus, we have a set {0}:={0,0}.

So, we don't need the definition of a set, we just need a pairing axiom...
 
  • #21
Ok, we know that the pairing axiom of ZF can construct 1={0}, but the pairing axiom doesn't state what a set is.

1 is a set because there exists a 0, such that 0 is a member of 1.
The pairing axiom states that 1={0}, but 1 is not a set because of the pairing axiom.

1 is a set because of the definition of set.
 
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  • #22
xxxx0xxxx said:
Ok, we know that the pairing axiom of ZF can construct 1={0}, but the pairing axiom doesn't state that {0} is a set.

Yes, it does. Because in ZF, everything is a set!
 
  • #23
But what then I ask, is a set?

Answer, "it is either something that has members or something that has no members with the special name 'the empty set'"
 
  • #24
A set is anything that satisfies the axioms of set theory. Like Hilbert said: "if the notions table and chair satisfy the axiom of set theory, then they are sets" (or something similar). You of course have the standard interpretation of set theory in which sets are simply "collection of objects", but it can have other interpretations, which are equally true. If one would only work under the standard intepretation, then there would be no forcing, no constructible universe, etc.
Choosing what interpretation of set theory one agrees to be standard is not mathematics, thus you will not find this, nor definitions of sets, in mathematics texts.
 
  • #25
But my definition can be found in many mathematics texts, Suppes 1960 for one.

Frege, BTW would criticize Hilbert's definition as failing to define set in terms of truth-functional statements, and only a statement about the consistency of set with the axioms of set theory.
 
  • #26
xxxx0xxxx said:
But my definition can be found in many mathematics texts, Suppes 1960 for one.

Suppes is a philosopher and not a mathematician.

Frege, BTW would criticize Hilbert's definition as failing to define set in terms of truth-functional statements, and only a statement about the consistency of set with the axioms of set theory.

Frege studied naive set theory and was ultimately proven wrong by Russel. It is Hilbert that tried to save set theory from contradictions. So Frege's criticisms don't really matter in my opinion.
 
  • #27
Well, if Suppes is merely a philosopher, then how can we accept Russell? Fortunately, Zermelo apparently discovered the paradox independently, since Russell (a philosopher and not a mathematician), somehow stumbled upon his eponymous paradox in an supreme accident of history. It seems though that Hilbert and Zermelo, being mathematicians, did not take it seriously enough to publish it, so a mere philosopher has usurped the honor.

I would say that Frege was proven no more wrong by Russell than Hilbert was proven wrong by Goedel.

But wait, I think Goedel was a philosopher, not a mathematician, so forget I said that.

But was Frege a mathematician or a philosopher, for if he were a mathematician, then Russell, being a philosopher could not have shown Frege to be wrong in any way.

Hopefully Frege was a philosopher so that the question is moot.

But now come to think, Goedel was thought only to be a mere logician, and no philosopher at all.

So what I said about Goedel showing that Hilbert was wrong could be true after all.

So at this point, Hilbert having ultimately been proven wrong by Goedel, could not have succeeded in saving set theory, and his attempts to do so don't really matter in my opinion.
 
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  • #28
xxxx0xxxx said:
Well, if Suppes is merely a philosopher, then how can we accept Russell?

There is a difference between doing mathematics and telling others how to do mathematics. Russell's paradox is mathematics, you can't deny it. But his entire principia mathematica is useless in my opinion.

I would say that Frege was proven no more wrong by Russell than Hilbert was proven wrong by Goedel.

Hilbert was not proven wrong by Gödel. Hilberts program was proven impossible by Gödel, that's something entirely different. That doesn't mean that Hilbert's point-of-view is wrong.

But wait, I think Goedel was a philosopher, not a mathematician, so forget I said that.

Gödel was a philosopher, but he was also a trained mathematician.

But was Frege a mathematician or a philosopher, for if he were a mathematician, then Russell, being a philosopher could not have shown Frege to be wrong in any way.

I never said that a philosopher can't prove a mathematician wrong. Don't twist my words please. But the point that we were discussing was which definitions and which axioms we should accept. In this case, I follow the mathematicians who actually do research about the subjects. Philosophers can come with other ideas, which can be equally correct, but if mathematicians do not follow their idea, then I also don't follow it. I never said philosophers are wrong, but it's up to the mathematical world to decide whether their contribution will be followed or not.

Hopefully Frege was a philosopher so that the question is moot.

He was a mathematician by training. But he also made significant contributions to philosophy.
 
  • #29
xxxx0xxxx said:
... Russell (a philosopher and not a mathematician)...
.
Actually, Russell was a trained mathematician. He studied mathematics at Trinity College, Cambridge.
 
  • #30
murshid_islam said:
.
Actually, Russell was a trained mathematician. He studied mathematics at Trinity College, Cambridge.

Thank God!

but,

A is STILL a set⇔∃z(z∈A∨A=∅)

This is valid whether we're talking ZF, NBG, New Foundations or the Bhagavad Gita ...
 
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  • #31
xxxx0xxxx said:
A is STILL a set⇔∃z(z∈A∨A=∅)

Why is "still" in all caps?
I personally do not understand why what you wrote is actually useful. A vector space is something that satisfies the vector space axioms. It's as simple as that. In today's mathematics a set is something that satisfies ZF. What more do you want?
 
  • #32
xxxx0xxxx said:
Thank God!

but,

A is STILL a set⇔∃z(z∈A∨A=∅)

This is valid whether we're talking ZF, NBG, New Foundations or the Bhagavad Gita ...

This is not true in NBG! Consider the class of all sets, this satisfies the right-hand side of your equivalence, but it's not a set.

It is true in ZF however, but iit's trivial since everything is a set there. Why don't you just accept the fact that mathematicians don't do things your way? I'm not saying it's wrong what you say, it's just not what we do.
 
  • #33
micromass said:
This is not true in NBG! Consider the class of all sets, this satisfies the right-hand side of your equivalence, but it's not a set.

It is true in ZF however, but it's trivial since everything is a set there. Why don't you just accept the fact that mathematicians don't do things your way? I'm not saying it's wrong what you say, it's just not what we do.

Ok rename set to class and the definition is valid... now

\mbox{A is a class} \Leftrightarrow \exists z( z \in A \vee A=\emptyset)

for NBG, then derive set from class, BTW now there is no class of classes, but get set "a" as a representation of class "A". All it is is renaming the model elements, only we use the word "class" for the word set. Now NBG is a model in ZF.

And... we can turn it around and make ZF a model in NBG.
 
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  • #34
pwsnafu said:
Why is "still" in all caps?
I personally do not understand why what you wrote is actually useful. A vector space is something that satisfies the vector space axioms. It's as simple as that. In today's mathematics a set is something that satisfies ZF. What more do you want?

Because, ZF is not the end all, be all of set theories, there are infinities of set theories.

ZF just happens to be the one that most people use; all the other set theories can be modeled in ZF and vice versa.
 
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  • #35
xxxx0xxxx said:
Ok rename set to class and the definition is valid... now

\mbox{A is a class} \Leftrightarrow \exists z( z \in A \vee A=\emptyset)

for NBG, then derive set from class, BTW now there is no class of classes, but get set "a" as a representation of class "A". All it is is renaming the model elements, only we use the word "class" for the word set. Now NBG is a model in ZF.

And... we can turn it around and make ZF a model in NBG.


A bit of a silly definition isn't it? Once your definition defines a set, and once a class. I don't like it. And I don't think many mathematicians do...

xxxx0xxxx said:
Because, ZF is not the end all, be all of set theories, there are infinities of set theories.

ZF just happens to be the one that most people use; all the other set theories can be modeled in ZF and vice versa.

Of course not all set theory can be modeled in ZF. You really need to open a set theory book...
 
  • #36
xxxx0xxxx said:
\mbox{A is a class} \Leftrightarrow \exists z( z \in A \vee A=\emptyset)
As an aside, in NBG, both sides of the equivalence in NBG are tautological predicates of one variable (A).
 
  • #37
Hurkyl said:
As an aside, in NBG, both sides of the equivalence in NBG are tautological predicates of one variable (A).

Yes, the convention is usually to drop the universal quantifier...

\forall A(\mbox{A is a class} \Leftrightarrow \exists z(z \in A \vee A = \emptyset))

All definitions are tautologies.
 
  • #38
xxxx0xxxx said:
All definitions are tautologies.
No, I mean, for example,
A is a class​
is a tautology.
 
  • #39
Hurkyl said:
No, I mean, for example,
A is a class​
is a tautology.

Well, that may be because you are not making a distinction between object language and metalanguage.
 
  • #40
There is no metalanguage in mathematics...
 
  • #41
micromass said:
There is no metalanguage in mathematics...

What he is probably referring to an axiomatic formalization of the logical inferences within set theory, known as a metatheory, in which one could e.g. formally say "\phi is unprovable" for a statement \phi in set theory, whereas set theory can only say "\phi" (or \neg \phi, or any combinations of conjunctions, negations, etc.. of set theoretical statements). A metalanguage consists of statements about statements in some axiomatic setting, like set theory. Such a metalanguage is necessary to be able to prove that certain things are unprovable in set theory, such as the continuum hypothesis, so called metatheorems. It is also used to prove that the axiom of choice is logically independent from ZF.
 
  • #42
xxxx0xxxx said:
Well, that may be because you are not making a distinction between object language and metalanguage.
What you wrote was a statement in the language of NBG, and you even stated it as "for NBG". If you mean something other than what you wrote, it's up to you to make that explicitly clear. It is not up to us to divine your intent, nor even to assume your being meaningful at all.
 
  • #43
Hurkyl said:
What you wrote was a statement in the language of NBG, and you even stated it as "for NBG". If you mean something other than what you wrote, it's up to you to make that explicitly clear. It is not up to us to divine your intent, nor even to assume your being meaningful at all.

Well, you need not divine much,

"A" is in the object language

"is a class" is a statement in the metalanguage.

"A is a class" is well-formed-formula by virtue of the definiens:

\exists z (z \in A \vee A = \emptyset)

which is composed of nothing but elements of the object language.
 
  • #44
disregardthat said:
What he is probably referring to an axiomatic formalization of the logical inferences within set theory, known as a metatheory, in which one could e.g. formally say "\phi is unprovable" for a statement \phi in set theory, whereas set theory can only say "\phi" (or \neg \phi, or any combinations of conjunctions, negations, etc.. of set theoretical statements). A metalanguage consists of statements about statements in some axiomatic setting, like set theory. Such a metalanguage is necessary to be able to prove that certain things are unprovable in set theory, such as the continuum hypothesis, so called metatheorems. It is also used to prove that the axiom of choice is logically independent from ZF.

Seems pretty straightforward to me.

ZF can serve as a metalanguage for NBG, or any other model.
 
  • #45
xxxx0xxxx said:
Seems pretty straightforward to me.

ZF can serve as a metalanguage for NBG, or any other model.

I'd love to see some kind of proof for that...
 
  • #46
xxxx0xxxx said:
"is a class" is a statement in the metalanguage.
"is a class" is a predicate in the language of NBG. (the trivially true one, to be precise)

I have no idea how you manage to read it as a "statement" "in the metalanguage" or what you could possibly mean by such a claim.

(maybe... you just aren't familiar with forms of first-order logic that offer a richer syntax for doing routine logical tasks?)




That's not quite true -- I was wondering if you were thinking of an interpretation of NBG, and you were using "class" to refer not to the type of that name in NBG (actually, I think it's usually shortened to Cls) but some metatype you're interpreting it into. But, of course, if you were doing that "A is a class" would make no sense if A was a variable in the language of NBG.
 
  • #47
micromass said:
I'd love to see some kind of proof for that...

Hurkyl said:
"is a class" is a predicate in the language of NBG. (the trivially true one, to be precise)

I have no idea how you manage to read it as a "statement" "in the metalanguage" or what you could possibly mean by such a claim.

(maybe... you just aren't familiar with forms of first-order logic that offer a richer syntax for doing routine logical tasks?)




That's not quite true -- I was wondering if you were thinking of an interpretation of NBG, and you were using "class" to refer not to the type of that name in NBG (actually, I think it's usually shortened to Cls) but some metatype you're interpreting it into. But, of course, if you were doing that "A is a class" would make no sense if A was a variable in the language of NBG.

See the attached...

Moderator's note -- removed copyrighted material[/color]
 
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  • #48
xxxx0xxxx said:
See the attached...

I don't see how that answers my question...
Firstly, your attachment assumes the existence of a inaccessible cardinal, which is a huge assumption. In fact it cannot be proven that ZFC+inaccessible cardinal is relatively consistent with ZFC. Thus it is an assumption I don't like to see.
Furthermore, a statement in ZFC is provable iff it is provable in NBG. Thus the assumption of a large cardinal seems to be a bit too much.

Also, you said that "any set theory can be phrased in terms of ZF". I have yet to see a proof for this. How would you phrase NF or MK in terms of ZF?
 
  • #49
micromass said:
I'd love to see some kind of proof for that...

See the attached...


http://matwbn.icm.edu.pl/ksiazki/fm/fm64/fm64126.pdf

Moderator's note: duplication of copyrighted material replaced with link to source[/color]
(article title: "Novak's result by Henkin's method")
 
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  • #50
That doesn't answer the question...
 

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