Proving the Empty Intersection of Intervals using Natural Numbers

[cragar]

Homework Statement

Prove that \bigcap_{n=0}^{\inf} (0,\frac{1}{n})=\emptyset

The Attempt at a Solution

since 0 is not included in our interval. eventually I will get to
(0,0) because I could pick a real as close to zero as I wanted and there would be a natural such that \frac{1}{n}<y therefore this intersection is empty.
but if my orginal intersection was [0,1/n] then this would not be empty.

 

[jbunniii]

You have the right idea, and you're correct that if (0,1/n) was replaced by [0,1/n], the intersection would not be empty. It would contain exactly one point: 0.

You could word your proof a bit precisely as follows:

Suppose the intersection is non-empty. Then there exists some x in the intersection:

x \in \bigcap_{n=1}^{\infty} (0, 1/n)

Since x is in the intersection, it means that x must be in every interval of the form (0, 1/n), and this is impossible because...

 

[cragar]

ok thanks for the input

 

[Deveno]

you might want to show that if x > 0, there exists k in N with 1/k < x, first.