# proving

• Oct 25th 2009, 04:28 AM
alexandrabel90
proving
how do you prove that U( n=1) ( -1/n, 1/n) ={0}?

first i tried to prove that RHS ⊂ LHS..
then to prove that LHS ⊂ RHS, i tried to do it in cases..

assume that x>0, then 1/x>0... so for 1/x ∈ ( -1/n, 1/n) , it means that 1/x < 1/n..then how do i prove by contraction if that is the case?

or is my method wrong?
• Oct 25th 2009, 05:35 AM
Jameson
Quote:

Originally Posted by alexandrabel90
how do you prove that U( n=1) ( -1/n, 1/n) ={0}?

first i tried to prove that RHS ⊂ LHS..
then to prove that LHS ⊂ RHS, i tried to do it in cases..

assume that x>0, then 1/x>0... so for 1/x ∈ ( -1/n, 1/n) , it means that 1/x < 1/n..then how do i prove by contraction if that is the case?

or is my method wrong?

Maybe it's just me, but I don't follow your initial statement. Are you using U as union? Can you explain your notation more please?
• Oct 25th 2009, 05:44 AM
alexandrabel90
yes. U is union where the LHS is the union of (-1/n , 1/n) from n=1 to n= infinity.
• Oct 25th 2009, 06:47 AM
Jameson
Quote:

Originally Posted by alexandrabel90
yes. U is union where the LHS is the union of (-1/n , 1/n) from n=1 to n= infinity.

Still don't get you. Do you mean (-1/n) U (1/n)?

What is weird about this question is that you're trying to prove the union is empty, which only is true if both sets are empty. Is it supposed to be the intersection instead?
• Oct 25th 2009, 06:54 AM
alexandrabel90
OHH sorry typo..

it should be that the INTERSECTION of all numbers from ( -1/n, 1/n) where n ranges from 1 to infinity consist of a set with the value 0 only.
• Oct 25th 2009, 06:55 AM
Bruno J.
What I believe she wants to show is

$\bigcap_{n=1}^\infty(-1/n, 1/n)=\{0\}$.

Am I right Alexandra? Because if the left-hand side is a union (as you wrote), then the right-and side should be $(-1,1)$.
• Oct 25th 2009, 06:57 AM
alexandrabel90
yes! that is what i wan to show..but i dont know how to show that
• Oct 25th 2009, 07:33 AM
alexandrabel90
sorry, but the question is (-1/n, 1/n), it means that the set consist of all elements that ranges from -1/n...0....1/n right? it does not mean that it has to be both negative and positive at the same time...
• Oct 25th 2009, 08:08 AM
HallsofIvy
No, there don't "have to be two sets". You can take the intersection of many sets!

This is the intersection of the open intervals (-1, 1), (-1/2, 1/2), (-1/3, 1/3), etc.

Alexandrabel90, RHS $\subset$ LHS should have been easy: 0 is in (-1/n, 1/n) for all n.

To prove LHS $\subset$ RHS, show that any x other than 0 is NOT in at least one of the intervals. If x is not 0 then it is either positive or negative. If x> 0, show that there exist some n such that 1/n< x (so x is not in (-1/n, 1/n) for that n). If x< 0, show that there exist some n such that x< -1/n (so x is not in (-1/n, 1/n) for that n.)

(Hint: Archimdean property)
• Oct 25th 2009, 08:27 AM
Jameson
Sigh, my mistake. Thank you HallsofIvy for helping me out.

"This is the intersection of the open intervals (-1, 1), (-1/2, 1/2), (-1/3, 1/3), etc."

This is the key point I wasn't understanding about the problem. You're right my wording was horrible about needing more than one set. Clearly subsets work. What I should have said was in what way intersections are being constructed isn't apparent to me.