A subsequence of a convergent sequence converges to the same limit. Questions on proof. (Abbott p 57 2.5.1)

Question

Solutions to Homework 3 doesn`t duplicate. We have to prove that if $(a_{n})$ is a sequence in $\mathbb{R}$ with $\displaystyle \lim_{n\rightarrow\infty} a_n =a$, and if $(a_{n_{k}})_{k\in \mathbb{N}+}$ is a subsequence of $(a_{n})$ , then $\displaystyle \lim_{k\rightarrow\infty}a_{n_{k}}=a$.

We first need to know that $n_{k}\geq k$ for all $k\in Z_{>0}$. This is proved by induction on $k$. I omit this.

Let $e >0$. $\displaystyle \lim_{n\rightarrow\infty}a_{n}=a$ is posited, so there's $\color{violet}{N}\in \mathbb{N}$ such that for all $n\in\mathbb{N}$, $n \ge N \implies |a_n-a| < e$.

Let $k\in \mathbb{N}$ with $k\geq N.$ Then ${n_k}\geq k\geq N$. Therefore $|a_{n_{k}}-a|<e$.

I know we must find $ N\in\mathbb{N}$ such that $\color{red}{n_k} \ge N\implies |a_\color{red}{n_k}-a| < e \quad (♫)$.
As the proof overhead shows, this $N$ is the same as the posited $\color{violet}{N}$. But what engenders $(♫)$ ?

Is proof saying $n \ge N \implies |a_n-a| < e$
implies $\color{red}{k} \ge N \implies |a_\color{red}{k}-a| < e$
implies $(♫)$, because $n_k \ge k \ge N$?

If yes, then I don't understand how $k$ can be replaced by $n_k$?

Answer 1

I'm going to post a separate answer because I think I can be clearer than the accepted answer.

I know we must find $ N\in\mathbb{N}$ such that $\color{red}{n_k} \ge N\implies |a_\color{red}{n_k}-a| < e \quad (♫)$.

No, this is not what you need to show! You need to show that there exists $N$, such that for all $\boldsymbol{k} \ge N, |a_{n_k} - a| < \epsilon$. This is because $\boldsymbol{a_{n_k}}$ is a sequence in $\boldsymbol{k}$, not a sequence in $\boldsymbol{n}$. It takes in an integer $\boldsymbol{k}$ and gives you an integer in return.

Is proof saying $n \ge N \implies |a_n-a| < e$
implies $\color{red}{k} \ge N \implies |a_\color{red}{k}-a| < e$
implies $(♫)$, because $n_k \ge k \ge N$?

Here is what the proof is saying. Fix any $k \ge N$. Then $n_k \ge k \ge N$. Thus $|a_{n_k} - a| < \epsilon$, Q.E.D.

If you have had trouble understanding any of the English in this answer, please let me know and I will try to clarify.

Answer 2

Right this is what is going on.

We first assume $ e\gt 0 $ is an arbitrary positive quantity. You've clearly understood that we need to produce a natural number $N$ such that for every element $a_{n_k}$ after the $N$th term of the sequence $(a_{n_k})$ we have $|a_{n_k} - a| \lt e$.

What we know is that (since $a_n \to a$) there is a a natural number $N$ such that every term in the sequence $(a_n)$ past the $N$th term is such that $|a_n - a| \lt e$. Now what is being argued is that if $a_{n_j}$ is a term in the subsequence $(a_{n_k})$ beyond "its" $N$th term then since $a_{n_j}$ is also a part of the original sequence and since $\{n_k\}$ is a strictly increasing sequence of natural numbers, $a_{n_j}$ is surely beyond the $N$th term in the original sequence and hence the inequality holds. This is what is projected by $ k \le n_k $.

Answer 3

The statement to be proven

For any sequence $(a_n)_{n \in \mathbb{N}}$ in $\mathbb{R}$ and $a \in \mathbb{R}$ such that $\lim_{n\to\infty} a_n = a$:

  For any sequence $(k_n)_{n \in \mathbb{N}}$ in $\mathbb{N}$ such that ( $k_{n+1} > k_n$ for any $n \in \mathbb{N}$ ):

    $\lim_{n\to\infty} a_{k_n} = a$.

Step 1

For any sequence $(a_n)_{n \in \mathbb{N}}$ in $\mathbb{R}$ and $a \in \mathbb{R}$ such that $\lim_{n\to\infty} a_n = a$:

  For any sequence $(k_n)_{n \in \mathbb{N}}$ in $\mathbb{N}$ such that ( $k_{n+1} > k_n$ for any $n \in \mathbb{N}$ ):

    [We simply expand the given condition by definition since we cannot use it directly.]

    For any $ε' > 0$:

      For some $m \in \mathbb{N}$:

        For any $n \in \mathbb{N}_{\ge m}$:

          $| a_n - a | < ε'$.

    ...

    For any $ε > 0$:

      For some $m \in \mathbb{N}$:

        For any $n \in \mathbb{N}_{\ge m}$:

          $| a_{k_n} - a | < ε$.

    [We need to get the above to get the next statement.]

    $\lim_{n\to\infty} a_{k_n} = a$.

Step 2

For any sequence $(a_n)_{n \in \mathbb{N}}$ in $\mathbb{R}$ and $a \in \mathbb{R}$ such that $\lim_{n\to\infty} a_n = a$:

  For any sequence $(k_n)_{n \in \mathbb{N}}$ in $\mathbb{N}$ such that ( $k_{n+1} > k_n$ for any $n \in \mathbb{N}$ ):

    For any $ε > 0$:

      [Create an $m$ given by the existential quantification using the desired $ε$.]

      Let $m \in \mathbb{N}$ such that:

        For any $n \in \mathbb{N}_{\ge m}$:

          $| a_n - a | < ε$.

      [I've renamed the "$m$" below to not conflict with the "$m$" given above.]

      ...

      [Now we have to show that at least one $c$ satisfies the statement under "for some $c$".]

      For some $c \in \mathbb{N}$:

        For any $n \in \mathbb{N}_{\ge c}$:

          $| a_{k_n} - a | < ε$.

    [We need to get the above to get the next statement.]

    $\lim_{n\to\infty} a_{k_n} = a$.

Step 3

For any sequence $(a_n)_{n \in \mathbb{N}}$ in $\mathbb{R}$ and $a \in \mathbb{R}$ such that $\lim_{n\to\infty} a_n = a$:

  For any sequence $(k_n)_{n \in \mathbb{N}}$ in $\mathbb{N}$ such that ( $k_{n+1} > k_n$ for any $n \in \mathbb{N}$ ):

    For any $ε > 0$:

      Let $m \in \mathbb{N}$ such that:

        For any $n \in \mathbb{N}_{\ge m}$:

          $| a_n - a | < ε$.

      [It turns out that this very $m$ satisfies the later statement we wanted!]

      Let $c = m$.

      [So it is enough to prove the following statement.]

      For any $n \in \mathbb{N}_{\ge c}$:

        ...

        $| a_{k_n} - a | < ε$.

    $\lim_{n\to\infty} a_{k_n} = a$.

Step 4

For any sequence $(a_n)_{n \in \mathbb{N}}$ in $\mathbb{R}$ and $a \in \mathbb{R}$ such that $\lim_{n\to\infty} a_n = a$:

  For any sequence $(k_n)_{n \in \mathbb{N}}$ in $\mathbb{N}$ such that ( $k_{n+1} > k_n$ for any $n \in \mathbb{N}$ ):

    [(1): The induction that you mentioned can be inserted here.]

    For any $ε > 0$:

      Let $m \in \mathbb{N}$ such that:

        (2): For any $n \in \mathbb{N}_{\ge m}$:

          $| a_n - a | < ε$.

      Let $c = m$.

      For any $n \in \mathbb{N}_{\ge c}$:

        $k_n \ge n$ [by (1)].

        Thus $k_n \ge c = m$. [This is why we chose $c$.]

        Thus $| a_{k_n} - a | < ε$ [by applying (2) to $k_n$].

    $\lim_{n\to\infty} a_{k_n} = a$.

Comments

See if the above makes sense. It is exactly how I constructed the proof from scratch. For it to work you must understand exactly what is meant by scoping and the logical rules. Most of it should be common sense once you get used to thinking in terms of nested contexts.

There are two ways of obtaining the above proof. The first is to blindly follow the logical rules to see what we can get. In simple cases like this it would actually work since all we needed to figure out was what $c$ would work inside there.

The second way is to understand the structure of limits of sequences. A limit means that for any tolerance level, when you go far enough along the sequence, eventually all subsequent terms will be near the limit within the chosen tolerance level. This tells you immediately that you want $c$ to be at least $m$ so that the subsequence has gone far enough.

The second way is of course the preferable method, but you still need to know the logical rules so that you can rigorously check that the intuitive solution is actually valid.