Proving x > 0 Implies Natural Number M for Xn > 0

[Hodgey8806]

Homework Statement

I'll try to restate it more clearer (I'm sorry I'm not good with using the more mathematical text type.

Prove that if lim (Xn) = x and if x>0, then there exists a natural number M such that Xn > 0 for all n≥M.

Homework Equations

Just to be clear, this limit definition is assumed to be to infinity as per our book for analysis.
(Xn) represents the sequence notation
And Xn will represent my term for a given n.

Scratch work pieces are that I want to choose an ε = x (the limit) at term XM
This would imply that for all n ≥ M, 0<Xn<2x

I would like to approach this using an ε-neighborhood.

The Attempt at a Solution

Assume that the lim(Xn) = x and x > 0.
=> Given any ε>0, |Xn-x| < ε
=> -ε + x < Xn < ε + x
=> There exists a natural number M s.t. 0 < XM < 2x (This is from the scratch work)
=> For all n ≥ M, 0<Xn<2x
=> Xn > 0 for all n ≥ M


I do want to make sure this is right. I realize that it is sloppy currently and I would love to have more help in writing this a bit nicer. It makes sense to me, but it's weird to choose a term that "fits" that epsilon.

Thanks for your help!

 

[LCKurtz]

OK, since your argument is in fact mathematically correct, I will show you how to better present it. My comments will be interspersed in red.

The Attempt at a Solution

Assume that the lim(Xn) = x and x > 0.

Start with: Pick $\epsilon = x$
=> [STRIKE]Given any ε>0, |Xn-x| < ε
[/STRIKE]

There is a natural number $M$ such that if $n > M$, $|X_n -x|<x$.

Now replace this statement:
=> -ε + x < Xn < ε + x
with: $-x < X_n - x < x$

Now if you add x to all three sides of that inequality you can delete the crossed out steps
[STRIKE]=> There exists a natural number M s.t. 0 < XM < 2x (This is from the scratch work)
=> For all n ≥ M,[/STRIKE] 0<Xn<2x
=> Xn > 0 for all n ≥ M

 

[Hodgey8806]

Thank you very much! Sorry for the late response. I just want you to know that the help is greatly appreciated, and the flow makes it much easier to read!