Math Is Fun Forum

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Check top

I got a very clumsy proof  , feel free to correct me , but don't go too hard on me , XD

Last edited by Stanley_Marsh (2007-05-17 09:23:53)

---

Numbers are the essence of the Universe

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Check top

If I am trying to prove than something exist in something , Can I design an algorithm to prove?

---

Numbers are the essence of the Universe

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Check top

What about just expanding on the definition of an open set?

A set E is open if for every x in E, there exists an epsilon > 0 such that the neighborhood of x with radius epsilon is a subset of E.

Let x be in E.  Then there exists an epsilon > 0 such that

.  So it must be that...

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Check top

It might be worthy to note that both rationals and irrationals are dense in R.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Check top

but I haven't learned about that , not yet . Then how should I solve that problem? Is finding an algorithm a correct way to do it?

---

Numbers are the essence of the Universe

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Check top

Is finding an algorithm a correct way to do it?

No, to show existence you need to expand on what an open set is.

You haven't learned about density?

Given any two real numbers a, b, with a != b, there exists a rational r such that a < r < b.

The same holds for irrationals.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Check top

A is dense in X if for any point x in X, any neighborhood of x contains at least one point from A.
Then the set of Rational and the set of irrational are dense in R^1 ,right? So any neighborhood of x in R^1 contain both rational and irrational .

But how to prove that the set of Rational and the set of  irrational are dense in R^1 ?

Last edited by Stanley_Marsh (2007-05-19 09:09:17)

---

Numbers are the essence of the Universe

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Check top

Given any two real numbers a, b, with a != b, there exists a rational r such that a < r < b.

so density basicly means that the set is not countable? i.e. integers are countable, and not dense, because there is no integer between two consecutive integers

reals are dense, there is always a real number between two real numbers, and are not countable.

Last edited by luca-deltodesco (2007-05-19 09:27:44)

---

The Beginning Of All Things To End.
The End Of All Things To Come.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Check top

No, because the rationals are countable and dense. Uncountability implies denseness though.
Edit: If I'm being really picky, perhaps I should say it implies 'not undenseness', because there are some sets where saying it is or isn't dense doesn't make sense.
Edit2: I stand corrected. It most usually does though. Ricky's counter-example was pretty much made up for the purpose of being a counter-example.

---

Why did the vector cross the road?
It wanted to be normal.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Check top

No, because the rationals are countable and dense. Uncountability implies denseness though.
Edit: If I'm being really picky, perhaps I should say it implies 'not undenseness', because there are some sets where saying it is or isn't dense doesn't make sense.

Before I go off on a tangent, take a look at this post.  There is an example of an uncountable set which is not dense, and on the real numbers no less.

Density only makes sense once the distance function is defined.  Let E be a subset of X.

Let d:ExE -> R (reals) be defined such that:

d(x, y) > 0 if x≠y
d(x, x) = 0
d(x, y) = d(y, x)
d(x, z) ≤ d(x, y) + d(y, z) for any y in E.

Then we call d the distance function.  A neighborhood of a point x is the set

A point x is a limit point of E if every neighborhood of x contains some point y such that y ≠ x and y is in E.

E is dense in X if everyone point of X is a limit point of E, or a point of E (or both).

So now think of this in terms of R being X, and Q being E.  The question becomes is every real number a limit point of the rationals?  This in turn becomes the question: Given any real number q and a radius r, can we always find a rational in N_r(q).  That question is answered on the real line by the property:

For every q and r, there exists a rational p such that q < p < q + r.

Since the distance function of the real line is just d(x, y) = |x - y| (absolute value).

[/tangent]

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."