Error in Solution of Topology Problem? 
Show that if $f : X \to X$ is a shrinking map and $X$ is a compact metric space, then $f$ has a unique fixed point. 

Here (on page 2) is a solution to the problem above, which I think contains an error. Let $A = \bigcap_{n=1}^\infty f^n(X)$. Here is the relevant part:

The crux of this argument is to show that $f(A) = A$. It is obvious that $f(A) \subseteq A$ by definition of $A = \bigcap_{n=1}^\infty$ (given $a \in a$, $a \in f^n(X)$ for all $n \in \Bbb{N}$ which implies $f(a) \in f^{n+1}(X)$ for all $n \in \Bbb{N}$.) 

Presumably, "$a \in a$" was intended to be "$a \in A$", but that is not the pertinent error. Certainly $f(a) \in f^{n+1}(X)$ for all $n \in \Bbb{N}$ is true, which translates to $f(a) \in \bigcap_{n=1}^\infty f^{n+1}(X)$. But $\bigcap_{n=1}^\infty f^{n+1}(X)$ is, prima facie, neither equal to, nor even a subset of, the set $\bigcap_{n=1}^\infty f^n(X)$, as $\{1,2\} \cap \{2,3\} \not\subseteq \{1,2\} \cap \{2,3\} \cap \{4\} = \emptyset$ suggests. So I don't see how $f(A) \subseteq A$ is obvious. 
 A: Except for the typo "$a\in a$", it is correct. However, it is not made explicit that the argument depends on the fact that we are not dealing with an arbitrary intersection but with the intersection of a nested sequence of sets. Since
$$f^{n+1}(X) = f^n\bigl(f(X)\bigr) \subseteq f^n(X)$$
for every $n\in \mathbb{N}$ it doesn't matter at what index the intersection starts (and we can even omit infinitely many terms, as long as we keep infinitely many), the intersection is always the same,
$$\bigcap_{n = 0}^{\infty} f^n(X) = \bigcap_{n = m}^{\infty} f^n(X)$$
for every $m\in \mathbb{N}$.
Then, using the fact that the image of an intersection is contained in the intersection of the images,
$$g\biggl(\bigcap_{i\in I} B_i\biggr) \subseteq \bigcap_{i\in I} g(B_i)$$
for all maps $g$ and families $\{B_i : i \in I\}$ of subsets of the domain of $g$, we obtain
$$f(A) = f\biggl(\bigcap_{n = 0}^{\infty} f^n(X)\biggr) \subseteq \bigcap_{n = 0}^{\infty} f\bigl(f^n(X)\bigr) = \bigcap_{n = 0}^{\infty} f^{n+1}(X) = \bigcap_{n = 1}^{\infty} f^n(X) = A.$$
A: You are right.  You also need to check that $f(a)\in f(X)$.  But this is obviously true.
