Real Analysis convergence proof

[elimenohpee]

Homework Statement

If the sequence xn ->a , and the sequence yn -> b , then xn - yn -> a - b

The Attempt at a Solution

Can someone check this proof? I'm aware you cannot subtract inequalities, but I tried to get around that where I indicated with the ** in the following proof. Does this make sense?

(for all epsilon > 0)(there exists a natural number K)(such that for all n > K) |(xn-yn) - (a-b)| < epsilon

|(xn - a) - (yn - b) | < epsilon
|(xn - a) + (-yn + b) | < epsilon **


choose 1.) |xn - a| < epsilon / 2
and 2.) |-yn + b| < epsilon / 2

1.) + 2.)

|(xn -a) + (-yn+b) | <= |xn - a| + |-yn +b| < epsilon

is this a valid proof?

 

[micromass]

All seems good, however:

and 2.) |-yn + b| < epsilon / 2

I think you still need to explain why you can choose an epsilon such that |yn+b|< epsilon/2.
It is indeed true that -yn converges to -b, but you still need to prove it! (even if it's trivial )

 

[elimenohpee]

All seems good, however:


I think you still need to explain why you can choose an epsilon such that |yn+b|< epsilon/2.
It is indeed true that -yn converges to -b, but you still need to prove it! (even if it's trivial )

great, thanks! and point taken lol :D

 

[SammyS]

(for all epsilon > 0)(there exists a natural number K)(such that for all n > K) |(xn-yn) - (a-b)| < epsilon

|(xn - a) - (yn - b) | < epsilon
|(xn - a) + (-yn + b) | < epsilon **


choose 1.) |xn - a| < epsilon / 2
and 2.) |-yn + b| < epsilon / 2

1.) + 2.)

|(xn -a) + (-yn+b) | <= |xn - a| + |-yn +b| < epsilon

is this a valid proof?

This the sort of thing you write on scratch paper to figure out a strategy for the proof you will submit for grading. What you have here assumes what you want to prove. It's sort of the reverse of your final proof.

The crucial step here is |(xn -a) + (-yn+b) | ≤ |xn - a| + |-yn +b| the triangle inequality.
And of course: |xn - a| + |-yn + b| = |xn - a| + |yn - b| .

To go forward,

Let ε > 0 .

Then since x_n converges there is some L that works for x_n, a and ε/2

Similarly, there is some M that works for y_n, b and ε/2​

Now, show how K is related to M & L to show that x_n - y_n converges to a - b .