My question is in the context of this answer.(https://math.stackexchange.com/a/3774433/710290). Please read this answer before getting to my question.
The statement of the theorem is
If $X$ is a complete, compact metric space and $f:X\to X$ is continous and satisfies $d(f(x),f(y))\lt d(x,y)$ for $x\neq y$ then the recursive sequence $f^{(n)}(x)$ is convergent.
Now, in the context of the answer, I have understood that $f$ has a unique fixed point but I can't understand what makes the recursive sequence convergent in the first place at all?
My thinking :
Let $\{a_n\}$ be the recursive sequence where $a_1=x$ and $a_{n+1}=f(a_n) ,\forall n\in
\mathbb{N}$. Then since the space $X$ is compact , there is a convergent subsequence $\{a_{r_n}\}$ .
Let $a_{r_n} \to l$ as $n\to \infty$.
Then $a_{r_n+1}=f(a_{r_n})\to f(l)$ as $n\to \infty$ by the continuity of $f$.
Similarly $a_{r_n+2}=f(a_{r_n+1})\to f(f(l))$ as $n\to \infty$
Similarly, for $k\in \mathbb{N}$
$a_{r_n+k}\to f^{(k)}(l)$ as $n\to \infty$
But I don't understand what to conclude from this. Please help.
As a sidenote, before you ask the question why didn't I ask there in the comments , I want to make it clear that I asked it there but didn't get any reply. So I thought of posting this as a separate question.
Thanks for your time and attention.
Best Answer
In the aforementioned link in the OP, it is proven that $f$ has a unique fixed point, say $w$.
To show that for any $x\in X$, $f^{(n)}(x)\xrightarrow{n\rightarrow\infty}w$, we show that any subsequence of $\{f^{(n)}(x)\}$ admits a subsequence that converges to $w$.
Following the notation of the link, define the function $Q(x):=d(f(x),x)$. Since $f$ continuous, so is $Q$; moreover, unless $x$ is a fixed point of $f$, we have that $$Q(f(x))=d(f(f(x)),f(x))<d(f(x),x)=Q(x)$$
If $Q(f^{(n)}(x))=0$ for some $n_0$, then $f^{(m)}(x)=f^{n_0}(x)$ for all $m\geq m_0$ and so, $f^{(n)}(x)\xrightarrow{n\rightarrow\infty}f^{(n_0)}(x)=w$ since $f^{(n_0)}(x)=f(f^{(n_0-1)}(x))=f^{(n_0-1)}(x)$.
Suppose $x$ such that $Q(f^{(n)}(x))>0$ for all $n$. Then, $$ \begin{align} Q(f^{(n)}(x))<Q(f^{(n-1)}(x))<\ldots<Q(x)\quad \forall n\in\mathbb{N}\tag{0}\label{zero} \end{align}$$ and so, $Q(f^{(n)}(x))$ converges. On the other hand, as $X$ is compact, any subsequence $\{f^{(n')}(x)\}$ of $\{f^{(n)}(x)\}$ admits a convergent subsequence $\{f^{(n_k)}(x)\}$. Say, $$f^{(n_k)}(x)\xrightarrow{k\rightarrow\infty}y\in X$$
For any $n$, there is a unique $k$ such that $n_k\leq n<n_{k+1}$; hence $$Q(f^{(n_{k+1})}(x))<Q(f^{(n)}(x))\leq Q(f^{(n_k)}(x))$$ and so, by the continuity of $Q$ $$\begin{align} \lim_nQ(f^{(n)}(x))=Q(y).\tag{1}\label{one} \end{align} $$ By $\eqref{zero}$, $$Q(f^{(n)}(x))>Q(y),\quad \forall n\in\mathbb{N}$$
We claim that $y$ is a fixed point. Otherwise, $Q(f(y))<Q(y)$. However, $Q(f(y))=\lim_k Q(f(f^{(n_k)}(x))\geq Q(y)$ which is a contradiction; hence $y$ is a fixed point, and by uniqueness $y=w$.
We have shown that any subsequence of $\{f^{(n)}(x)\}$ admits a subsequence that converges to the unique fixed point $w$ of $f$. From this, we conclude that in fact $f^{(n)}(x)\xrightarrow{n\rightarrow\infty}w$.
Edit: This is to address a comment from the OP:
Lemma: Suppose $(X,d)$ is a metric space, $a\in X$ and $\{a_n:n\in\mathbb{N}\}\subset X$. The sequence $a_n$ converges to $a$ iff any subsequence $a_{n'}$ of $a_n$ admits a subsequence $a_{n''}$ that converges to $a$.
Here is a short proof:
($\Longrightarrow$) Obvious.
($\Longleftarrow$) Suppose $a_n$ does not converge to $a$. Then, there is $\varepsilon>0$ such that for any $k\in\mathbb{N}$, there is $n_k\in \mathbb{N}$ such that $d(a_{n_k},a)\geq \varepsilon$. Without loss of generality, we may assume that $n_k<n_{k+1}$. Then $\{a_{n_k}:k\in\mathbb{N}\}$ is a subsequence of $\{a_n:n\in\mathbb{N}\}$, and no subsequence of $\{a_{n_k}\}$ converges to $a$ (for $d(a_{n_k},a)\geq\varepsilon$ for all $k$).