# Is it called a closed set?

1. Jul 30, 2014

### johann1301

If you take two arbitrary numbers from a set N - lets say N stands for the natural numbers - and add them together, the sum will always be an element of N.

In my language, there is a word for this, but i dont know what it is in english? If i translate it from norwegian, it would be something like;

"The set N is closed for addition"

Is "closed" the correct word?

2. Jul 30, 2014

### micromass

Yes, closed is the right word. Other word is stable.

3. Jul 30, 2014

### johann1301

You mean "stable" is a synonym for "closed"? (in this mathematical context)

4. Jul 30, 2014

### micromass

Yes.

5. Jul 30, 2014

### johann1301

Then i have a couple more questions...

I have heard that a set A is called a group if A is closed for both addition and subtraction, its called a ring if its altso closed for multiplication, and finally, its a body if its closed for them all(division as well).

Are the words "group", "ring" and "body" the correct english words to describe a set?

Last edited: Jul 30, 2014
6. Jul 30, 2014

### micromass

Group and ring are correct. A body is commonly used in many languages, but not english. In english it's called a field (or if you don't demand multiplication to be commutative; a division ring).

7. Jul 30, 2014

### johann1301

Last question then... (i think)

I was given the problem to prove that the real numbers R is a field. The only way i can think of how to do this is by showing that √-1 does not show up in the calculation. Would this be a sufficient proof?

8. Jul 30, 2014

### micromass

What would your definition of $\mathbb{R}$ be?

9. Jul 30, 2014

### johann1301

I dont have a mathematical/algebraic definition, but i would say its any number, as long its not an imaginary number. Or at least the coefficient of i, would be zero; a+bi=a+0i=a+0=a.

it could be;

zero
pie
tau
√2
2
-3
e

but not i...

10. Jul 30, 2014

### WWGD

There is a list of axioms used to describe a field. You then only need to show these axioms are satisfied.

EDIT :Sorry, this was supposed to be a reply to Johann about showing the Reals are a field.

Last edited: Jul 30, 2014
11. Jul 30, 2014

### Matterwave

A linguistic tidbit: In English we usually say a set is closed under addition or subtraction or multiplication or what have you, and not closed for addition, etc.

12. Jul 30, 2014

### micromass

If you don't have an actual definition, then it's impossible to prove it. Maybe they don't expect you to prove it but just to give some vague argument for why it could be true?

13. Jul 30, 2014

### Fredrik

Staff Emeritus
My favorite definition of $\mathbb R$ is of the form "a field such that..." I we use that definition, there's nothing to prove. There's a similar definition of $\mathbb C$, I mean a definition that says that $\mathbb C$ is a field right at the start. This definition too makes the problem almost too easy. All you would have to do is to show that the sum and the product of two arbitrary real numbers is real. Everything else follows from the definitions that say that $\mathbb R$ is a subset of $\mathbb C$ and that $\mathbb C$ is a field.

The problem becomes much more interesting and difficult if you use one of the more explicit definitions of $\mathbb R$, for example the Dedekind-cut definition, or the equivalence classes of Cauchy sequences definition.

I think you really need to find out what definition of $\mathbb R$ you're supposed to use.

14. Jul 30, 2014

### FactChecker

No. Don't introduce complex numbers. That is extraneous. Instead, you can define the reals by their decimal representation. Then define addition, subtraction, multiplication, division by the usual algorithms. Prove that the definitions of a field are satisfied.

15. Jul 30, 2014

### micromass

You're in for a world of pain if you're going to define the real numbers by their decimal representations. It's really not a fun proof at all.

16. Jul 30, 2014

### FactChecker

I agree that it would be bad. But I can't think of another basic definition without defining it as the topological closure of rationals. And I think that would be a whole new can of worms. Are there any other options?

P.S. Even using the decimal representation, I don't know how to rigorously define addition of two numbers that require carrying digits from infinitely small ( like 5.555555..... + 5.5555555......) So I guess I give up. Maybe the limits of rationals is the only rigorous way to do it.

17. Jul 31, 2014

### micromass

No, you're right. There is no easy definition. The decimal representation definition is definitely the most "intuitive", but is very difficult to work with.

It must somehow be possible to define it. I have no idea how though. And additionally, you have this awkward problem that you must define 1 = 0.9999... for some reason to make everything work. This is quite annoying.

18. Jul 31, 2014

### FactChecker

I looked through my abstract algebra books and didn't see any that addressed this. I think it would require defining reals as limits of rationals. I see that your initial response, "What would your definition of R be?", got right to this issue. I didn't appreciate how crucial that was at the time.

19. Jul 31, 2014

### johann1301

"R = The set of all real numbers, that is, all the numbers on the numberline."

This is the only definition wich is given.

Im starting to think maybe this is the case:

Due to what all of you are saying, it seems like a very complicated problem to solve. Way to difficult for a beginner in set-theory.

20. Jul 31, 2014

### johann1301

(i think this thread might be more approriate i the homework section though, sorry about that)