How can I prove the nested intervals theorem using a proof by contradiction?

[irony of truth]

If I_n = (0, 1/n) where n is any natural no. is a sequence of nested intervals, then the intersection of all the I_n is empty.

I was able to get the proof similar as that of above problem but the I_n is closed, that is, I_n = [0, 1/n]... and further, the intersection of all the I_n is {0}.


What I am trying to do is to prove by contradiction because it is the easiest way to prove it. How do I construct this?

 

[lurflurf]

If I_n = (0, 1/n) where n is any natural no. is a sequence of nested intervals, then the intersection of all the I_n is empty.

I was able to get the proof similar as that of above problem but the I_n is closed, that is, I_n = [0, 1/n]... and further, the intersection of all the I_n is {0}.


What I am trying to do is to prove by contradiction because it is the easiest way to prove it. How do I construct this?

let x be in the intersection
0<x
x<1/n all natural numbers n -> x=<0
x=<0 & x>0 -><-

 

[irony of truth]

lurflurf,

I apologize for asking this... I will "repeat" what you mentioned because I am a bit "slow"...

If I were to prove it by contradiction, will I assume that x is the intersection...?
Then, from the nested interval property (is it?), 0 < x < 1/n for all natural no. n.

From here on...you mentioned that "-> x=<0
x=<0 & x>0 -> <-"... do you mind if you can explain to me?

 

[matt grime]

by the definitions of the real numbers there is no real number that is strictly positive and less than 1/n for all n, that's all.

 

[HallsofIvy]

More "primitively", if x is any positive real number then 1/x exists and is a positive real number. By the "Archimedean property", there exist an integer n> 1/x so 1/n< x.