# Homework Help: Advanced calculus proof

1. Sep 12, 2008

### emira

1. The problem statement, all variables and given/known data

Prove that: ⋂_(n=1)^∞▒〖(0,1/n]=∅〗

2. Relevant equations

If tried with mathematical induction, we would need to know the procedure follows:
This would be my P(n) statement: ⋂_(n=1)^∞▒〖(0,1/n]=∅〗

now we would need to find out if P(1) is true and then assume P(k) for some k$$\geq$$1, is true. Then we would need to prove the statement P(k+1) is also true.

3. The attempt at a solution

I tried using induction, but our professor, nor the book explained any examples that actually go to infinity, all we covered and proved with induction has been up to "n".

Thus my thinking consists of taking the semiintervals:
a) (0,1/1]∩(0,1/2]=(0,1/2]
so it continues for even smaller semiintervals, and the intersection between a bigger semi-interval with a larger semi-interval is the smaller of the two...up to the point of having

(0,1/(huge nr close to ∞)]∩(0,1/∞]=(0,1/∞]=(0,0]

The only thing I cannot explain is since this semi-interval contains zero, how can it be an empty set? So if someone could give me an idea or a hint, anything at all, I would appreciate it a lot!

emira

2. Sep 12, 2008

### Dick

What you want to prove is that for any x in the reals, x is not in the intersection of (0,1/n] for all n. So all you have to do is given x, find an n such that (0,1/n] does not contain x.

3. Sep 12, 2008

### emira

What does x as a real number have to do with it though? what we have to prove is that the infinite intersection of this semi-intervals that get smaller and smaller, is an empty set.
plus how could i prove that x is not an element of the small interval, and how does that help me into proving what i have to prove?

4. Sep 12, 2008

### Dick

If x<=0 it isn't in any of the intervals, so it's not in the intersection. If x>0 can you show there is an n such that 1/n<x? If so then it isn't in (0,1/n] so it's not in the intersection. I assumed you were working over the reals. The rationals are ok too. Pick an integer n greater than 1/x. If n>1/x then x>1/n, right?

Last edited: Sep 12, 2008