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 :
and
or and
or
Showing that y is a point of closure
To show now that y is a point of accumulation we must additionaly show that
How do we show that??
.
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 .
The is a rational number &
The is a rational number &
If the is a rational number &
Now the sequence
So &
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.