# The set {} and the set {0}

1. Nov 6, 2008

### SW VandeCarr

The set {} the set {0} differ in that the latter contains one member. Yet they both express the null value. Why is having two nonequal sets expressing the null value not a contradiction?

2. Nov 6, 2008

### HallsofIvy

Staff Emeritus
First you are going to have to say what you mean by "the null value"! {} is the empty set, {0} is not. How do they both express "the null value"?

3. Nov 7, 2008

### CRGreathouse

I'm going to question the premises. {0} does not "express the null value", so there's no problem. (Actually, I don't know why it would be a problem for the predicate "{this} expresses the null value" to be true for two different sets, but that's a different matter.)

4. Nov 7, 2008

### SW VandeCarr

In various texts the empty set is referred to as the 'null set' and the zero vector as the 'null vector'. In the definitions of the positive integers in terms of sets (Peano and others), {}=0 and {0,{0}}={{}, {{}}}=1. The null vector in R^n is 0(1),...,0(n). If {}=0 then we should be able to write the null vector as {}(1),...,{}(n). However if there is only one empty set, its iteration within a single expression for the zero vector would seem to be a contradiction. You should not be able to count n empty sets.

5. Nov 7, 2008

### Tac-Tics

You are using "iteration" here where it is inappropriate, and you should mind what you're actually doing. Remember that vectors are n-tuples of numbers. And a tuple is NOT the same as a set, and 0 = (0, 0, ..., 0) = ({}, {}, ..., {}) is NOT the same as {0, 0, ..., 0} = {0} = {{}}

The Peano construction of the integers (and similar constructions, say formulating functions or relations as sets) is only there to show that the integers arise from (an otherwise number-less) set theory. Other than that, they are totally useless! An integer isn't a set. A function isn't a set. A tuple isn't a set. These are only there to help in the total axiomatizing of a formal system, and don't help the human mathematician in solving any useful problems outside of that field.

6. Nov 7, 2008

### SW VandeCarr

Yes, an n-tuple is not a set. I'm interested in foundations and set theory is all about foundations. The statement {}=0 relates a set to a number. The n-tuple is an ordered series of numbers. If I substitute {} for each zero as the above relation allows me to do, I get an ordered series of countable sets. Since there is only one empty set, I would like to know why this is not a contradiction.

Last edited: Nov 7, 2008
7. Nov 7, 2008

### Tac-Tics

Where did the idea of a series come in? Tuples are not series either.

The standard way to define a pair in set theory is this:

(x, y) = {{x}, {x, y}}

From there, you define an n-tuple as:
(x1, x2, ..., xn) = (x1, (x2, ..., xn)), if n > 2
(x1, x2), if n = 2

So if you had a 3-tuple of zeros, expanding it by these rules, we would get:
0 = (0, 0, 0) = (0, (0, 0)) = (0, {{0}, {0, 0}}) = {{0}, {0, {{0}, {0, 0}}}} = {{{}}, {{}, {{{}}, {{}, {}}}}}.

This is very clearly distinct from both 0 = {} and {0} = {{}}. It's also a good example of why these constructions are only theoretical tools. You show it can be done, and you continue pretending tuples and integers are primitive constructions in your formal system.

8. Nov 8, 2008

### SW VandeCarr

With the exception of my informal use of the term "series", you've not challenged anything I've stated in post 4. I'm not saying the positive integers are primitive constructions and it's not my formal system. I'm citing the Peano construction of zero and the positive integers from set theory. In this construction {}=0. If so, the ordered pair (0,0) can then be written as ({},{}). I'm saying that this is suspect since it uses a unique set to express two non-interchangable components of the null or zero vector in R^2. How can we have countably many representations of the empty set in a single mathematical object? What does two empty sets mean when there is only one empty set? I agree that a set is not a number. That is just my point.

Regarding {0}, what is the cardinality of this set? What is the cardinality of {}? If they are both of same cardinality and {}=0, why can't we say that {0}=0?

Last edited: Nov 8, 2008
9. Nov 8, 2008

### Hurkyl

Staff Emeritus
Problem 1: you're confusing two homonyms -- two different things that have the same name. The additive identity of a vector space has nothing to do with the empty set, despite the fact the symbol '0' is commonly used to refer to both of them.

Problem 2: you haven't understood the notion of a model
This is not a definition of the positive integers in terms of sets. Two different things are being done simultaneously here:

Thing 1: this is a method for assigning certain labels to certain sets. This has nothing to do with the natural numbers, aside from the fact its using the same set of symbols to construct the labels.

Thing 2: this is (part of) the construction of a model of the natural numbers. It is setting up a correspondence between
1. The symbols used in the language of the natural numbers
2. Certain sets and functions
And then after this interpretation is constructed, it will be shown that if you interpret Peano's axioms by replacing all of the number-theoretic expressions with the corresponding set-theoretic expressions, they will become theorems of set theory.

One.

Zero.

Because they don't contain the same elements. Specifically:
$$0 \in \{ 0 \}$$
$$0 \notin 0$$

Last edited: Nov 8, 2008
10. Nov 9, 2008

### InvisibleBlue

-Regarding your problem with ({}(0), {}(1),...,{}(n),...) - {}(0), {}(1), etc are not different empty sets. They are all {}. Th integers assigned to them only determine the position. It's ok to do this with sets, in the same way that it is ok to have n-tuple (0,0,0,...) in a ring where 0 in a unique element. In ({}(0), {}(1),...,{}(n),...), {}(0) and {}(1) are different elements of the tuple, but not because they are different sets, merely because they are in different positions. Now if you swap the position of a few of these {}(i) you get the same thing since:
({}(0), {}(1),...,{}(n)) = (0(0), 0(1),...,0(n)) = (0(1), 0(0),...,0(n)) = ({}(1), {}(0),...,{}(n))
as you know two n-tuples (a(0), a(1),...), (b(0), b(1),...) are the same if a(i) = b(i) for all i. So if you continue permuting the elements of ({}(0), {}(1),...,{}(n)) as above you get that {}(0) = {}(1) = {}(2) =...= {}(n).

-Regarding your original question about {} and {0}. The cardinality of finite sets is defined to be the number of elements in them. And if {0} confuses you, then think of the function
f: {0} -> {p} by f(0) = p . Here p is any object you. The function f creates a bijection between {0} and {p}. You can now treat {0} and {p} the same. Their cardinalities for example are the same.

11. Nov 9, 2008

### SW VandeCarr

-Regarding your original question about {} and {0}. The cardinality of finite sets is defined to be the number of elements in them. And if {0} confuses you, then think of the function
f: {0} -> {p} by f(0) = p . Here p is any object you. The function f creates a bijection between {0} and {p}. You can now treat {0} and {p} the same. Their cardinalities for example are the same.

So if we have f:{0} -> {p} and f(0)=p, can I have p (not {p}) be the empty set, or am I violating some rule of set theory? If p is the empty set, then {p} is {{}} and f(0)={}. Is this "legal"?

12. Nov 9, 2008

### InvisibleBlue

Yes there is nothing illegal about it. The empty set is also an object (as is any other set). But you should remember that in the context of above {} is being treated as an object not a set, otherwise there is room for confusion:
The range of f as defined above is not empty, it has one element in it and that is the object {} which is also called the empty set. This should not be confused with a "function" whose range is empty, which can be argued not to be a function (by definition of function)

So you should not think of f(0)={} as "f(0) is empty" as this would suggest that f does not map 0 to anything. You should think of it as "f(0) is the object 'empty set'".

13. Nov 11, 2008

### SW VandeCarr

Can we also write the inverse f({})=0? Also, is f(0)={} consistent with Hurkyl's informative post (9) in that it sets up a correspondence between sets (as a model) and the natural numbers (including 0)?

Last edited: Nov 11, 2008
14. Nov 11, 2008

### enigmahunter

$$f:\varnothing \rightarrow A$$ exists since for every x in the domain $$\varnothing$$ there is a unique y in the codomain A such that $$(x,y) \in \varnothing$$ . It is a vacuously truth since there are not any x in the domain.

No $$f:A \rightarrow \varnothing$$ exists by definition of a function since no element y in $$\varnothing$$can be chosen.

$$f:A \rightarrow B$$
If f is a function and $$\varnothing$$ is an element of A (A can be a collection of sets), 0 is an element of B, $$f(\varnothing) = 0$$ can be chosen.

If f is a function and 0 is an element of A, $$\varnothing$$ is an element of B, $$f(0)=\varnothing$$ can be chosen.

Last edited: Nov 11, 2008
15. Nov 11, 2008

### Coin

$$f^{-1}$$({}) would be = 0 if we imagine f to be Hurkyl's correspondence.

But f({}) would be undefined. This is because what Hurkyl describes is a model of the natural numbers, not a model of "the natural numbers plus set theory"; there is no way to model {} inside the model, you would have to invent a more elaborate model for that.

What you are trying to do is treat things that exist inside and outside the model as if they're interchangeable; you can't do this. Statements about the model only remain true as long as you follow the model's rules.

16. Nov 11, 2008

### enigmahunter

A model of the Peano axioms for natural numbers is a triple (N,0,S), where N is an infinite set and 0 belongs to N, S is a successor function.

$$S(a) = a\cup\{a\}$$
$$0=\varnothing$$
$$1=s(0)=\varnothing \cup \{ \varnothing \} = \{ \varnothing \}=\{0\}$$
$$2=\{0,1\}$$
.....

If you use the above model for the Peano axioms, there is only one function defined in the structure, which is a successor fuction. The model gives you an interpretation for this function. If you use a different kind of function, it would be a different structure and it may not be a model for the Peano axioms any more.

In general cases, I think $$f(\varnothing)=0$$ can be defined as I mentioned in the previous post.

Last edited: Nov 11, 2008
17. Nov 11, 2008

### Hurkyl

Staff Emeritus
I guess I should explain why one would ever make a model of the natural numbers. There are two main points:

(1) Relative consistency of Peano's axioms
The fact we can construct a model of the natural numbers tells is that if Peano's axioms are inconsistent, then so is (Zermel-Fraenkel) set theory. Contrapositively, if set theory is consistent, then Peano's axioms are consistent too.

(2) Applying natural numbers to set theory
We know a lot about the natural numbers. When we construct a model of the natural numbers in set theory, it allows us to apply that knowledge to the study of sets.