Is the Proof Valid for the Convergence of the Sequence x_n?

[bedi]

Homework Statement

Let x_n be a convergent sequence with a ≤ x_n for every n, where a is any number. Prove that a ≤ lim x_n when n→∞.

Homework Equations

Definition of limit. The usual ε, N stuff.

The Attempt at a Solution

Let lim x_n = x and choose ε=x_n-a. Hence we have |x_n - x| < x_n - a which shows that -x<-a and thus x>a.

Is this valid?

 

[micromass]

Homework Statement

Let x_n be a convergent sequence with a ≤ x_n for every n, where a is any number. Prove that a ≤ lim x_n when n→∞.

Homework Equations

Definition of limit. The usual ε, N stuff.

The Attempt at a Solution

Let lim x_n = x and choose ε=x_n-a.

You do realize that $\varepsilon$ depends on n?? For a different n, you'll get a different $\epsilon$. Do you really want that??
Also, why is $\varepsilon>0$? Specifically, why is it nonzero?

 

[tiny-tim]

hi bedi!

(try using the X₂ button just above the Reply box )

Let lim x_n = x and choose ε=x_n-a

but ε has to be independent of n

(ooh, micromass beat me to it! )

 

[bedi]

Yes, you are right. However I still can't see the solution :(

 

[tiny-tim]

try assuming the contrary

 

[bedi]

Alright, so this will imply that x_n converges to both a and x which is a contradiction. Am I right?

 

[micromass]

Alright, so this will imply that x_n converges to both a and x which is a contradiction. Am I right?

Take $x_n=2$ for all n and take $a=0$. Then certainly $x_n$ does not converge to a. So no, you're not right.

 

[bedi]

But I assumed that a>x. So -a<-x and x_n-a<x_n-x<ε. Hence x_n-a<ε ?

 

[micromass]

But I assumed that a>x. So -a<-x and x_n-a<x_n-x<ε. Hence x_n-a<ε ?

Although the idea is there, the proof is still not very nice. For example, what is $\varepsilon$?? You got to say things like this, not just introduce them without telling anybody what it is.

 

[tiny-tim]

But I assumed that a>x …

… what is $\varepsilon$?? You got to say things like this, not just introduce them without telling anybody what it is.

bedi, you know this is supposed to be a delta,epsilon proof …

so you must define delta, and you must define epsilon​