$f$ has countably many fixed points .Yes/No

Question

Consider the function $f : (0, \infty) \to \mathbb{R} $ defined by $f(x) = \sin \frac{1}{x}$ , for all $ x \in (0, \infty)$ .

Pick correct statement(s) from the following

$1.$ $f$ has countably many fixed points.

$2.$ For any $a >0$ , the restriction map $ f |_{(a, \infty) } : (a, \infty) \to \mathbb{R} $ has infinitely many fixed points.

$3.$The restriction map $f |_{(0,1]}:(0,1] \to \mathbb{R} $ has finitely many fixed points.

$4.$ The restriction map $ f |_{[1, \infty) } : [1, \infty) \to \mathbb{R} $ has no fixed points.

My attempts:

Both options $1$ and $2$ are correct in my opinion because $\sin \frac{1}{x}$ intersect $x$ at countably infinite points.

Am i right ?

Answer 1

No, you are not. Note that $\sin(1/x)$ oscillates infinitely many times in the interval $(0,\infty)$ but it only does so finitely many times in any interval bounded away from zero. In particular, if $a>1$, the restriction in question never has a fixed point. (While there are infinitely many intersections, you missed that where they are matters too!)

Hence, $(2)$ is certainly not correct, while $(4)$ is. Both are obvious from the graph:

$(1)$ merits a proper argument of such. While you do know infinitely many fixed points exist, how do you know if it is a countable number? (After all, $f(x) = x$ trivially has uncountably many fixed points.)

Answer 2

PrincessEev's answer is helpful, but not complete because as she mentions (1) requires a proper argument. Here is one showing countable fixed points:

You want to find cardinality of the set of solutions of $\sin t = \frac{1}{t}$ where I put $t = \frac{1}{x}$. Now suppose I label each half wave by a number i.e. Consider the sets $\cup_n A_n$ where $A_n = (n\pi, (n+1)\pi]$. Clearly the number of $A_n$ are countable, and you can formally show that the number of solutions of $\sin t = \frac{1}{t}$ lying in $A_n$ for any $n$ is atmost $2$, hence the number of fixed points is countable.