Sequence and subsequence - real analysis

[Dassinia]

Hello,
Solving last exam and stuck in this exercise

Homework Statement

Consider an increasing sequence {x_n} . We suppose ∃ x∈ℝ and {x_nk} a sebsequence of {x_n} and x_nk→x
a/ Show that for any n∈ℕ , ∃ k∈ℕ as n≤n_k
b/ Show that x_n→x

Homework Equations

3. The Attempt at a Solution [/B]
For b/ it is easy.
But for a/ I really don't know how to do that

thanks

 

[Stephen Tashi]

∃ k∈ℕ as n≤n_k

I don't know how to interpret that statement. What is it that happens "as" $n \le n_k$ ?

 

[HallsofIvy]

If you mean "n∈ℕ , ∃ k∈ℕ such that n≤n_k" that just says that, given any integer n, there exist an "n_k", an index from the subsequence, larger than n. And that comes from the fact that the subsequence is infinite.

 

[Dassinia]

Yes sorry it is such that , it was late !
I don't know where to start from to get to this result ?

 

[SammyS]

Yes sorry it is such that , it was late !
I don't know where to start from to get to this result ?

Well, what do you mean by n_k ?

Isn't {n_k} an increasing sequence in ℕ , so that {x_nk} is a subsequence ?