# 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 $$m_k (assuming affirmative on question 1b because the domain is an infinite subset of $$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.