Confused about basics of subsequences Hello I am a bit confused in regard to subsequences. The following image is taken from Introduction to Real Analysis, by Bartle and Sherbert.
For example, why is the following true?
My question mostly is , why can we say that $$n_{k} \ge k$$ ? for example what if $n_{2}=1$? what if the sequence was the harmonic for example and we took our subsequence to be every other term for example.
So what is meant by this, and why is saying it all is for natural numbers. Are we not dealing with regular ones anymore?
I hope someone can  see where my confusion is and help to clarify for me, thanks
For example what if we had a sequence $(a_{n})=0.1,0.01,0.001,0.0001,…$
and a subsequence $b_{2n}=0.001,0.00001,…$
then I don't get what it is saying for example b_{2} is of course not greater then 2.
So i think there is a really deep misunderstanding for me, possibly an example could help as well
 A: Let $(x_{n})$ be a sequence (without loss of generality, let $n$ run through $1,2,\dots$); then a subsequence of $(x_{n})$ is by definition obtained from a strictly increasing map $k \mapsto n_{k}: \{ 1,2,\dots \} \to \{ 1,2,\dots \}$ and is denoted by $(x_{n_{k}})$.
Let $(x_{n_{k}})$ be a subsequence of $(x_{n})$. If $k=1$, then $n_{k} = n_{1} \geq 1$; if $k \geq 1$ and $n_{k} \geq k$, then $n_{k+1} > n_{k} \geq k$, implying that $n_{k+1} \geq k+1$; by mathematical induction we have proved that for every $k \geq 1$ we have $n_{k} \geq k$.
Note that it is $n_{k} \geq k$ rather than $x_{n_{k}} \geq k$.
A: Edit. You are confusing in the sequence $b_{2n}$. The statement no says that $b_2\ge 2$. Since we have $b_{2n}$, the statement says that $2n\ge n$ which is true.
The statement $n_k\ge k$ is about the subindexes $n_k$ and $k$ of the sequences $(a_{n_k})$ and $(a_k)$.
Note that the key in the definition of a subsequence is that:


*

*we have a sequence $(a_n)_{n=1}^\infty$;

*and we have a sequence $(b_m)_{m=1}^\infty$;

*the sequence $(b_m)_{m=1}^\infty$ is a subsequence of the sequence $(a_n)_{n=1}^\infty$ if there exists a function $f\colon\Bbb N\to\Bbb N$ which is strictly increasing (i.e., $f(n+1)>f(n)$ for all $n\in\Bbb N$) such that $$b_n = a_{f(n)} \text{ for all } n\in\Bbb N;$$

*the function $f$ define a sequence (note that a sequence is just a function from $\Bbb N$ to $\Bbb N$);

*we  are using the notation $(n_k)$ to represent the function $f$. When one write $a_{n_k}$ is the same that $a_{f(k)}$ (i.e., we have $n_k=f(k)$);

*finally, the statemente $n_k\ge k$ is equivalent to $f(k)\ge k$.


Remember that $f$ is strictly increasing, so if we start $f(1)=x$, the clearly $f(2)>x$ and $f(3)>f(2)>x$, and so on.
Thus, the minimum number what we can to start $f(1)$ is $f(1)=1$; similarly, the minimum number assigned to $f(2)$ is $f(2)=2$, etc. So it is no possible that $f(k)<k$.

You can prove by induction that for every natural number $k$, we have $n_k\ge k$.
We cannot have $n_2=1$ What is the value of $n_1$ in this case? Obviusly cannot be $n_1=1$, beacuse $1=n_1<n_2=1$, a contradiction. By other hand, we cannot have $n_1>1$ because $n_k$ is increasing (note that $n_2=1<n_1$ leads a contradiction).

By definition (see below), we have a sequence $(n_k)_{k=1}^\infty$ of natural number. Also, it is increasing. So basically we have a increasing function $f:\Bbb N\to\Bbb N$ (with $f(k)=n_k$) and we need to show that $f(k)\ge k$ for every natural number $k$. (Note by definition of increasing function that $f(k)<f(k+1)$ for every $k$ in $\Bbb N$.)
We prove this by induction. For $k=1$, clearly we have $f(1)\ge1$ since $f(1)\in\Bbb N=\{1,2,\dotsc\}$. Now, suppose inductively that $f(k)\ge k$, we need to show that $f(k+1)\ge k+1$.
Thus, we have $f(k+1)>f(k)\ge k$, i.e., $f(k+1)>k$. Since $k$ is a natural number, we obtain $f(k+1)\ge k+1$. (Note that $a<b$ implies $a+1\le b$ when $a,b\in\Bbb N$.)

Definition of subsequences. Let $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ be sequences. We say that $(b_n)_{n=1}^\infty$ is a subsequence of $(a_n)_{n=1}^\infty$ iff there exists a function $f\colon\Bbb N\to\Bbb N$ which is strictly increasing (i.e., $f(n+1)>f(n)$ for all $n\in\Bbb N$) such that $$b_n = a_{f(n)} \text{ for all } n\in\Bbb N.$$
A: I think that you have misunderstood what a sub-sequence is, which is what seems to be causing this confusion. Let the sequence be represented by the function f: N to X, where X is any non-empty set. Let g be a function (n(1), n(2), ...), an infinite subset of natural numbers, to N, such that g(k) = k. Then, a sub-sequence is the function h, the composition of g and f, i.e., gof. Now, since domain of h is an ordered set, n(1) < n(2) < n(3) <... . 
Now, notice that in the proof, they do not say that the elements of the range, x(n(k)) >= k, but that n(k) >=k. This is easy to intuitively see, and we can show it by induction. Since {n(1), n(2), ...} is a subset of natural numbers, and so, 1<= n(1) < n(2) ,... . Now, consider n(2) > n(1), so since n(2) is a natural number, n(2) >= n(1) + 1 >= 2.
Let this hold for some natural k, that is, n(k) >= k. Now, n(k+1)> n(k), so n(k+1) >= n(k) + 1 >= k + 1. Thus, by the Principle of Mathematical Induction, n(t) >= t for all t in N.
Moreover, in your example, you have said n(2) = 1, which is not possible, since n(1) >=1 (1 is the least natural number), and 1<= n(1) < n(2) = 1, which implies that 1<= n(1) <1, which is  contradiction. Thus, your counter-example is erroneous.     
