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Here is Prob. 7, Sec. 28, in the book Topology by James R. Munkres, 2nd edition:

Let $(X, d)$ be a metric space. If $f$ satisfies the condition $$ d\big( f(x), f(y) \big) < d(x, y) $$ for all $x, y \in X$ with $x \neq y$, then $f$ is called a shrinking map. If there is a number $\alpha < 1$ such that $$ d \big( f(x), f(y) \big) \leq \alpha d(x, y) $$ for all $x, y \in X$, then $f$ is called a contraction. A fixed point of $f$ is a point $x$ such that $f(x) = x$.

(a) If $f$ is a contraction and $X$ is compact, show $f$ has a unique fixed point. [Hint: Define $f^1 = f$ and $f^{n+1} = f \circ f^n$. Consider the intersection $A$ of the sets $A_n = f^n(X)$.]

(b) Show more generally that if $f$ is a shrinking map and $X$ is compact, then $f$ has a unique fixed point. [Hint: Let $A$ be as before. Given $x \in A$, choose $x_n$ so that $x = f^{n+1}\left(x_n\right)$. If $a$ is the limit of some subsequence of the sequence $y_n = f^n \left( x_n \right)$, show that $a \in A$ and $f(a) = x$. Conclude that $A = f(A)$, so that $\mathrm{diam}\, A = 0$.]

(c) Let $X = [0, 1]$. Show that $f(x) = x - x^2/2$ maps $X$ into $X$ and is a shrinking map that is not a contraction. [Hint: Use the mean-value theorem of calculus.]

(d) The result in (a) holds if $X$ is a complete metric space, such as $\mathbb{R}$; see the exercises of \Sec. 43. The result in (b) does not: Show that the map $f \colon \mathbb{R} \to \mathbb{R}$ given by $f(x) = \left[ x + \left( x^2 + 1 \right)^{1/2} \right]/2$ is a shrinking map that is not a contraction and has no fixed point.

Here is my MSE post on Prob. 7 (a).

Here I'll only be attempting a solution to Prob. 7 (b).

My Attempt

Prob. 7 (b):

Here is another Math SE post on this very problem. However, here I'll attempt a proof using the hint offered by Munkres.

We first show that the shrinking map $f$ is uniformly continuous on $X$. Given a real number $\varepsilon > 0$, let us choose a real number $\delta$ so that $0 < \delta \leq \varepsilon$. Then for all $x, y \in X$ for which $d(x, y) < \delta$, we would obtain $$ d \big( f(x), f(y) \big) \leq d( x, y) < \delta \leq \varepsilon.$$ Since $\varepsilon > 0$ was arbitrary, it follows that $f$ is uniformly continuous on $X$.

Let $i_X \colon X \to X$ denote the identity map on $X$, defined by $$ i_X (x) \colon= x \ \mbox{ for all } \ x \in X. \tag{Def. 0} $$ Now let us put $$ f^n \colon= \begin{cases} i_X \ & \mbox{ if } n = 0, \\ f \circ f^{n-1} \ & \mbox{ if } n = 1, 2, 3, \ldots. \end{cases} \tag{Def. 1} $$ Next, let us put $$ A_n \colon= \begin{cases} X \ & \mbox{ if } n = 0, \\ f^n(X) \ & \mbox{ if } n = 1, 2, 3, \ldots. \end{cases} \tag{Def. 2} $$ Then we find that, for each natural number $n$, $$ A_n = f \left( A_{n-1} \right). \tag{0} $$

Now as the maps $i_X$ and $f$ are both continuous mappings of the compact space $X$ into itself, so are all the maps $f^n$ in (Def. 1) above, and thus all the sets $A_n$ in (Def. 2) above are all compact subspaces of $X$; moreover since $X$, being a metric space, is a Hausdorff space and since each set $A_n$ is a compact subspace of $X$, each set $A_n$ is also closed in $X$. And, as each set $A_n$ is closed in $X$, so is the intersection of these sets. Let us put $$ A \colon= \bigcap_{n=0}^\infty A_n. \tag{Def. 3}$$ Then as $A$ is a closed set in the compact space $X$, so $A$ is also compact (as a subspace of $X$).

As $f$ is a mapping of set $X$ into itself, so we have $f(X) \subset X$, that is, $$A_1 \subset A_0.$$ Now suppose that, for some natural number $k$, we have $$ A_k \subset A_{k-1}. $$ Then using (0) above we find that $$ A_{k+1} = f \left( A_k \right) \subset f \left( A_{k-1} \right) = A_k.$$ Therefore by induction we can conclude that $$ A_n \subset A_{n-1} \ \mbox{ for } n = 1, 2, 3, \ldots. \tag{1} $$

Thus $\left\{ \ A_n \ \colon \ n = 0, 1, 2, \ldots \ \right\}$ is a nested sequence of non-empty closed sets in the compact space $X$; therefore their intersection is non-empty, that is, set $A$ in (Def. 3) above is non-empty.

We now show that $\mathrm{diam}\, X$ is finite. Let $p$ be any point of $X$. Then the collection $$ \left\{ \ B_d \left(p, N \right) \ \colon \ N \in \mathbb{N} \ \right\},$$ where $$ B_d \left( p; N \right) \colon= \{ \ x \in X \ \colon \ d(x, p) < N \ \},$$ forms an open covering of the compact space $X$; so some finite sub-collection of this collection also covers $X$; that is, there exist finitely many natural numbers $N_1, \ldots, N_n$ such that the collection $$ \left\{ \ B_d \left(p, N_1 \right), \ldots, B_d \left(p, N_n \right) \ \right\}$$ of open balls covers $X$. Let $$ M \colon= \max\left\{ \ N_1, \ldots, N_n \ \right\}. $$ Then we obtain $$ X = B_d (p, M).$$ Thus for any points $x, y \in X$, we have $$ d(x, y) \leq d(x, p) + d(p, y) < M + M = 2M.$$ So $$ \mathrm{diam}\, X \leq 2M < +\infty. $$ Thus we have shown that $$ \mathrm{diam}\, X < +\infty. \tag{2} $$ Hence from (Def. 3) above and from (1) we can also conclude that $$ \mathrm{diam}\, A \leq \mathrm{diam}\, A_n \leq \mathrm{diam}\, A_{n-1} < +\infty \ \mbox{ for } n = 1, 2, 3, \ldots. \tag{3} $$

Now suppose $x \in A$. Then $x$ is in each set $A_n = f^n(X)$, and so there exists a point $x_n \in X$ such that $x = f^{n+1}\left(x_n\right)$ for each $n = 1, 2, 3, \ldots$; let us put $$ y_n \colon= f^n\left(x_n\right) \ \mbox{ for each } n = 1, 2, 3, \ldots. \tag{Def. 4} $$ Then $\left( y_n \right)_{n \in \mathbb{N}}$ being a sequence in the compact metric space $(X, d)$ has a convergent subsequence; let $\left( y_{\varphi(n)} \right)_{n \in \mathbb{N}}$ be this subsequence for some strictly increasing function $\varphi \colon \mathbb{N} \to \mathbb{N}$, and let $a$ be the limit of this sequence.

What next? How to proceed from here?

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  • $\begingroup$ Isn't a shrinking map also a contraction? $\endgroup$ – Aniruddha Deshmukh Mar 8 at 3:56
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    $\begingroup$ @AniruddhaDeshmukh No, not necessarily. $\endgroup$ – Henno Brandsma Mar 8 at 6:46
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We have $y_{\varphi(n)} \in A_{\varphi(n)} \subset A_{\varphi(m)}$ for $n \ge m$. Hence $a = \lim y_{\varphi(n)} \in A_{\varphi(m)}$ because $A_{\varphi(m)}$ is closed. This implies that $a \in \bigcap_m A_{\varphi(m)} = A$. Since $f$ is continuous and $y_{\varphi(n)} \to a$, we get $f(y_{\varphi(n)}) \to f(a)$. But the sequence $f(y_{\varphi(n)}) = f^{\varphi(n)+1}(x_{\varphi(n)}) = x$ is constant and we conclude $f(a) = x$.

This shows $A \subset f(A)$.

Assume that $d = \text{diam} A > 0$. Then we find sequences $(x_n), (y_n)$ in $A$ such that $d(x_n,y_n) \to d$. Since $A$ is compact, we may w.l.o.g. assume that both sequence converge to points $x, y \in A$. We get $d(x,y) = d$. Choose $a, b \in A$ such that $f(a) = x, f(b) = y$. Then $d = d(x,y) = d(f(a),f(b)) < d(a,b)$, which contradicts the definition of $d$.

Therefore $\text{diam} A = 0$ which is possible only when $A$ contains a single point $a$. This is a fixed point of $f$. Since $A$ trivially contains all fixed points of $f$, we are done.

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