Math Beauties in need of HELP

1. Oct 6, 2007

marlen

Math Beauties in need of HELP!!!!

1. The problem statement, all variables and given/known data

the question is....why the set of irrational numbers is not an open subset of Real numbers?

2. Relevant equations

none that i am aware of

3. The attempt at a solution

can we say that.....
assume that the irrational numbers are open. but we know that the set of real numbers is infinite and much larger than the set of irrational numbers. therefore, there is not a one to one mapping of irrational numbers to rational numbers.

2. Oct 6, 2007

arildno

Well, make a tiny open interval around an irrational number.

Does that interval only contain irrationals?

3. Oct 6, 2007

d_leet

In the reals what is the complement of the irrational numbers? Is this set open or closed?

4. Oct 6, 2007

HallsofIvy

Staff Emeritus
Being open or not has nothing to do with cardinality. Instead look at the definition of "open set".

Do you know this: given any two real numbers, x< y, there exist an irrational number z such that x< z< y. If p is any rational number rational number, then for any $\delta> 0$ there exist an irrational number in $(p-\delta , p+\delta )$. Is p an interior point of the set of rational numbers? What does that tell you about whether the set of rational numbers is open or not?

That's just what arildno said a few seconds ahead of me!

5. Oct 6, 2007

marlen

No, the interval will contain either another irrational number or a rational number.

6. Oct 6, 2007

arildno

120, to be exact, Halls!

7. Oct 6, 2007

arildno

Is such a thing allowable if the set of irrationals were to be open?

8. Oct 6, 2007

marlen

The complement of irrational numbers is the set of rational numbers. and this set can't be closed or open, i think...because the real numbers is an open set and i know that the union of two open sets is open. but irrational numbers are not open, so i'm stuck

9. Oct 6, 2007

arildno

Well, but the union of two non-open sets can perfectly well be open, can't it?

10. Oct 6, 2007

marlen

no. ... so what you are saying is that if we have an irrational number x, and for some small r>0 we get the interval (x-r, x+r) which contains either a rational or irrational number, meaning we cannot draw an open-ball around our x, meaning that the set of irrationals cannot be open.

11. Oct 6, 2007

marlen

i suppose but does that matter

12. Oct 6, 2007

arildno

You were the one in the preceding post who was "stuck" on precisely that issue.
You are, however, right, so there is no reason for you to remain stuck on it, since it happens to be irrelevant.

13. Oct 6, 2007

arildno

Precisely!

Q.E.D.

14. Oct 6, 2007

marlen

I'm not really sure how to answer this question.

15. Oct 6, 2007

marlen

but we can't just assume that there lies a rational or irrational number in the the interval of (x-r, x+r) can we?

16. Oct 6, 2007

arildno

For every r>0, the open r-interval about the irrational number will contain (infinitely many) rationals.

Therefore, the irrational set cannot be open.

17. Oct 6, 2007

marlen

Yea, but you can't just say that. I know it's true but how do i show that.

18. Oct 6, 2007

arildno

We certainly can:

Remember, any REAL number can adequately be given, say, a decimal representation.

Now, pick an irrational, regard its decimal expansion.

Now, truncate that infinite sequence of decimals into a finite sequence, and you get a RATIONAL number.

Thus, you may construct a sequence of rationals CONVERGING upon the irrational, i.e, there will be rationals arbitrarily close to any, and every, irrational number.

19. Oct 6, 2007

marlen

Thank you arildno and everyone else who helped us. It makes sense now! ;)

20. Oct 7, 2007

nebbish

Actually the important thing is that every interval of the reals contains a rational number. Therefore every interval around every irrational point contains a point that is not in the set of irrationals. Therefore the irrationals are not an open set.

For the irrationals to be open there would need to be an interval around each irrational containing just irrationals. But there isn't. Most likely your professor or your text has a proof showing that each interval of the reals contains a rational and therefore the irrationals are not open.