Proving the Convergence of Sequence a_n to a: How to Show Unique Partial Limits?

[MathematicalPhysicist]

i have that lim x_n=lim y_n=a
and we have the sequence (x1,y1,x2,y2,...)
i need to show that this sequence (let's call it a_n) converges to a.

well in order to prove it i know that if a_n has a unique partial limit then it converges to it to it, but how do i show here that it has a unique partial limit?
i mean if we take the subsequences in the even places or the odd places then obviously those subsequences converges to the same a, but i need to show this is true for every subsequence, how to do it?
thanks in advance.

 

[HallsofIvy]

i have that lim x_n=lim y_n=a
and we have the sequence (x1,y1,x2,y2,...)
i need to show that this sequence (let's call it a_n) converges to a.

well in order to prove it i know that if a_n has a unique partial limit then it converges to it to it, but how do i show here that it has a unique partial limit?
i mean if we take the subsequences in the even places or the odd places then obviously those subsequences converges to the same a, but i need to show this is true for every subsequence, how to do it?
thanks in advance.

I think you are making too much of this. Since x_n converges to a, given $\epsilon> 0$ there exist N₁ such that if n> N₁ then $|x_n-a|< \epsilon$. Since y_n converges to a, given $\epsilon> 0$ there exist N₂ such that if n> N₁ then $|y_n-a|< \epsilon$.

What happens if you take N= max(N₁, N₂)?

 

[MathematicalPhysicist]

yes i see your point.
if we take the max of the indexes then from there, we have that either way a_n equals x_n or y_n, and thus also a_n converges to a, cause from the maximum of the indexes both the inequalities are applied and thus also |a_n-a|<e, right?

 

[HallsofIvy]

Yes, that is correct.