# Subsequence - is it a mapping of the original sequence or the sequence of elements?

As part of my attempts to study calculus I've stumbled upon a conceptual question in the area of subsequences.

Take the following two sequences:

1. An = 1, 1/2, 1/3, 1/4, ..., 1/n
2. Bn = 1/2, 1/4, ..., 1/n


And assume that Bn is a subsequence of An. To my understanding a Subsequence is defined to be a sequence such that every number in Bn appears in An, and every appearance of a such number in An is in a lower index than the next appearance.

On the other hand, a Subsequence is defined to be a function that takes natural numbers and returns a new index in the original sequence. If to schematically draw it:

    N       N
bn ---> an ---> R


This shows us that Bn receives a neutral number and returns an index in An which later returns the value of the sequence in that same N.

What bothers me is the following:

By the last definition Bn is obviously not a subsequence as all its elements aren't a natural number in the first place. However, the mapping { 2,4,6,...,2n } are indexes and they really do point to the relevant numbers in An. If that's the case, is { 2,4,6,...,2n } the real subsequence, while { 1/2, 1/4, ..., 1/n } is some representation of that subsequence?

Thanks!

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This may not be exactly what you're looking for, but I'm suspicious of your function definition of a subsequence that you mention. One can view a sequence $(a_n)$ as a function $f: \mathbb{N} \to \mathbb{R}$ by $f(n) = a_n$. This does not require that the numbers $a_n$ be natural! The idea is that for each natural number $n$ (the index) we associate a number $a_n$ ( which we call an element of the sequence). How do we get a subsequence from a sequence? The idea of a subsequence is that we choose a strictly monotone function $g: \mathbb{N} \to \mathbb{N}$, and then set $h = f \circ g$ and define $b_n = h(n)$ The values in the sequence are the $b_n$ values, which are real numbers. Then, the subsequence is given by the function $h$, and for each natural number $n$ (the index) we have $b_n = h(n) = a_{g(n)}$. So, the values of $b_n$ are just values of the original sequence $a_n$, but chosen by possibly skipping some elements in $a_n$. Note that since $g$ is monotone, it is one-to-one and increasing, so if $n < m$, then $g(n) < g(m)$, so our subsequence is always going in the "right direction"; that is the subsequence won't hop backwards and take values of $a_n$ before $a_{g_n}$. Since $g(n)$ is always increasing, it's possible that $g(1) = 2$ and $g(2) = 200$, so $b_1 = a_2$ and $b_2 = a_{200}$, so lots of members of $a_n$ can get skipped in the subsequence $b_n$.
Note that this aligns with your first definition. The members of the subsequence $b_n$ are just members of the sequence $(a_n)$, but the members in $(b_n)$ are occurring at a possibly lower index value than they do in $a_n$ (because we are possibly skipping elements of $a_n$).
So, for your example, $f(n) = \frac{1}{n} =a_n$. For your subsequence, we only want to grab every other element of $a_n$. That is, the index should skip and only grab the even indices. To do this, we let $g(n) = 2n$,and define $h(n) = f \circ g(n) = \frac{1}{2n}$.
If you don't like this "function perspective", the alternative view is the following. Let $(a_n)$ be a sequence and let $n_k$ be a sequence of natural numbers that is increasing. That is, $n_{k} < n_{k+1}$. Then, we define the subsequence $b_k$ by $b_k = a_{n_{k}}$.