A tiny topological claim in a larger proof of mine

[Jamin2112]

Homework Statement

I'm writing a proof for my Real Analysis III class, and in one clause I claim that the intersection of my countably infinite set of intervals {E_n} where E_n=(1+1/2+1/3+1/4+...+1/n , ∞), has the property that the infinite intersection of all E_n's equals ∅ (This would be a lot easier to explain if I finally took the time to learn Latek).

Homework Equations

Not many.

The Attempt at a Solution

Obviously, since ∑1/k ---> ∞, my intersection of my set of intervals gets infinitely smaller in length, ultimately approaching (∞, ∞)=∅. I'd like someone to explain this in formal topological language since I'm not taking another topology class this summer.

 

[Fredrik]

Two sets are equal if and only if they have the same members, so you need to prove that this intersection has no members. Let x be an arbitrary real number. Can you prove that x is not a member of this intersection?

The LaTeX guide for the forum can get you started with LaTeX very quickly.

 

[Jamin2112]

Let x be an arbitrary real number. Can you prove that x is not a member of this intersection?

Obviously. Since Ʃ(1/k)=1+1/2+1/3+...+1/n ---> ∞, for any x in (1+1/2+1/3+...+1/n, ∞) there is an N≥n large enough that 1+1/2+1/3+...+1/N ≥ x, and consequently (1+1/2+1/3+...+1/n, ∞) & (1+1/2+1/3+...+1/N, ∞) = (1+1/2+1/3+...+1/N, ∞) does not contain contain x.

 

[Fredrik]

Right, there's an n such that $x\notin\big(\sum_{k=1}^n\frac 1 k,\infty\big)$. (Not sure why you started with an x in that interval though. That makes the proof a bit weird. Just let x be an arbitrary real number). Do you also understand why this implies that $x\notin\bigcap_{m=1}^\infty\big(\sum_{k=1}^m \frac 1 k,\infty\big)$, and why that implies that $\bigcap_{m=1}^\infty\big(\sum_{k=1}^m \frac 1 k,\infty\big)=\emptyset$?