real analysis – Show that a subsequence is a subset of the original sequence

My main question is the 4th point, but I hope you can clarify some things for me along the way.

The definition of a sequence says that a function $ a: mathbb {N} to S $ is a sequence on a set $ S $, denoted $ (a_n) $.

  1. Can i freely restrict the area of ​​function $ a $ and still call it a sequence? In particular, a) is it valid to define $ a_n $ on a finite subset of $ mathbb {N} $ b) on an infinite subset of $ mathbb {N} $?

Let's say that at each term in the sequence $ (a_n) _ {n in mathbb {N}} $, I have to define a new sequence from there. I first define a sequence $ (m_k) $ who maps $ {k in mathbb {N}: k geq n } to mathbb {N}, forall n in mathbb {N} $ with $ m_k <m_ {k + 1} $ (assuming affirmative on question 1b because the domain is an infinite subset of $ mathbb {N} $). Then, like $ (a_ {m_k}) $ is the composition of the sequence $ (a_n) $ and the increasing sequence $ (m_k) $, by definition $ (a_ {m_k}) $ is a subsequence of $ (a_n) $.

  1. Is it a good way to show that these new sequences $ (a_ {m_k}) $ are subsequences?

A set of points defined for the sequence $ (a_n) $ East $ left {a_n: n in mathbb {N} right } $.

  1. How do you define a set of points for a subsequence $ (a_ {m_k}) $? East $ forall n in mathbb {N}, left {a_ {m_k}: k geq n right } $ well?

Assuming yes on question 3, the main question is:

  1. How to prove this $ forall n in mathbb {N}, left {a_ {m_k}: k geq n right } subseteq left {a_n: n in mathbb {N} right } $? In other words, that the set of points defined for a subsequence is a subset of the set of points in the original sequence. I think i should take a term off $ left {a_ {m_k}: k geq n right } $ and deduce that its also in the whole $ left {a_n: n in mathbb {N} right } $, but I don't know how to do it rigorously.

To give you context on my questions, I want to show that $ sup { left {a_n: n in mathbb {N} right }} geq sup { left {a_ {k}: k geq n right }} $. Given that $ (a_n) $ is bounded, defines $ left {a_n: n in mathbb {N} right } $ and $ left {a_ {k}: k geq n right } $ are delimited. Having $ left {a_ {k}: k geq n right } subseteq left {a_n: n in mathbb {N} right } $ I would prove the supremums as in the question Prove the supremum of a subset is smaller than the supremum of the whole.