# Thread: prove Q" =R

1. ## prove Q" =R

Prove that all the accumulation points of the rationals are the Real Nos.

We know from a theorem in Analysis that:

Given a real No x and ε>0 there exists a rational No y such that :

$|x-y|<\epsilon$and $y\in Q$

or $y\in (x-\epsilon,x+\epsilon)$ and $y\in Q$

or $(x-\epsilon,x+\epsilon)\cap Q\neq\emptyset$

Showing that y is a point of closure

To show now that y is a point of accumulation we must additionaly show that

$y\neq x$

How do we show that??

2. Originally Posted by alexandros
Prove that all the accumulation points of the rationals are the Real Nos. How do we show that??
Between any two numbers there is a rational number.

$\left( {\exists r} \right)\left[ {r \in \mathbb{Q} \cap \left( {x - \varepsilon ,x} \right)} \right]$ so $r\not= x$.

3. Originally Posted by Plato
Between any two numbers there is a rational number.

$\left( {\exists r} \right)\left[ {r \in \mathbb{Q} \cap \left( {x - \varepsilon ,x} \right)} \right]$ so $r\not= x$.
But given r and x if x-ε<r for all ε>0 ,then $x\leq r$ .

Suppose the opposite and x>r ,then put ε = x-r ,and x-(x-r)<r => 0<0 :contradiction

Hence x can be equal to r

4. Originally Posted by alexandros
But given r and x if x-ε<r for all ε>0 ,then $x\leq r$ .
Suppose the opposite and x>r ,then put ε = x-r ,and x-(x-r)<r => 0<0 :contradiction
Hence x can be equal to r
You are seriously confusing two different concepts.
If $a~\&~b$ are any two real numbers then there is a rational number between them.

The proper translation of 'between' implies that it is distinct from both.

5. Originally Posted by Plato
You are seriously confusing two different concepts.
If $a~\&~b$ are any two real numbers then there is a rational number between them.

The proper translation of 'between' implies that it is distinct from both.
.

You are not listening to what i am saying and writing.

PLease read the whole thread again ,if you are really want to help and not just force your ideas whether wrong or right.

Start with any theorem you like and write a complete proof showing Q' = R,IF YOU wish to do so of course.

Writing parts of the proof make me more confuse.

The versions of the theorem concerning reals and rationals or irrationals that i know are:

1)Given any two girls a<b ,there exist a rational or irrational y, such that: a<y<b

2) Given any real x ,then for any ε>0 there exist a rational or irrational ,y such that

|x-y|<ε.

None of them proves that : Q' =R

6. Originally Posted by alexandros
.You are not listening to what i am saying and writing.
PLease read the whole thread again ,if you are really want to help and not just force your ideas whether wrong or right.

Start with any theorem you like and write a complete proof showing Q' = R,IF YOU wish to do so of course.
Writing parts of the proof make me more confuse.
The versions of the theorem concerning reals and rationals or irrationals that i know are:
1)Given any two girls a<b ,there exist a rational or irrational y, such that: a<y<b
2) Given any real x ,then for any ε>0 there exist a rational or irrational ,y such that
|x-y|<ε.
None of them proves that : Q' =R
I beg your pardon.
I taught this material for forty years.
I know exactly what you are saying and how really confused you are.

Every real number is a limit point of the set of rational numbers.
Here is the proof. Suppose that $\gamma\in\mathbb{R}$.
The is a rational number $r_1\in (\gamma -1, \gamma )$ & $r_1>\gamma -1~\&~r_1\not=\gamma$
The is a rational number $r_2\in (\max\{\gamma-\frac{1}{2},r_1\}, \gamma )$ & $r_2>r_1~\&~r_2\not=\gamma$

If $n\ge 3$ the is a rational number $r_n\in (\max\{\gamma-\frac{1}{n},r_{n-1}\}, \gamma )$ & $r_n> r_{n-1}~\&~r_n\not=\gamma$

Now the sequence $\left( {r_n } \right) \to \gamma$
So $\gamma\in\mathbb{Q}^{\prime}$ & $\mathbb{Q}^{\prime} =\mathbb{R}$

7. Originally Posted by Plato
I beg your pardon.
I taught this material for forty years.
I know exactly what you are saying and how really confused you are.

Every real number is a limit point of the set of rational numbers.
Here is the proof. Suppose that $\gamma\in\mathbb{R}$.
The is a rational number $r_1\in (\gamma -1, \gamma )$ & $r_1>\gamma -1~\&~r_1\not=\gamma$
The is a rational number $r_2\in (\max\{\gamma-\frac{1}{2},r_1\}, \gamma )$ & $r_2>r_1~\&~r_2\not=\gamma$

If $n\ge 3$ the is a rational number $r_n\in (\max\{\gamma-\frac{1}{n},r_{n-1}\}, \gamma )$ & $r_n> r_{n-1}~\&~r_n\not=\gamma$

Now the sequence $\left( {r_n } \right) \to \gamma$
So $\gamma\in\mathbb{Q}^{\prime}$ & $\mathbb{Q}^{\prime} =\mathbb{R}$
I wonder what axiom or theorem supports such sequence construction.

8. Originally Posted by alexandros
I wonder what axiom or theorem supports such sequence construction.
Do you understand that between any two numbers there is a rational number?
That is a theorem.

If $R\in\mathbb{R}$ then there is a rational number $r_1$ such that $R-1.

Thus there is a rational number $r_2$ such that $r_1.

Keep the up.

9. Originally Posted by Plato

Keep the up.
.

There is no theorem or axiom to justify the " Keep the up" process uppon which the constraction of your sequence is based.

10. Alexandros: Please don't badger the contributors; we are fortunate to have their input and rely on their help. If you don't like what someone has to say, just move on. In my opinion, Plato seems very experienced and provided thoughtful input to you before the thread went personal.