Graph of Sine of Reciprocal

Theorem

Let $G$ be the graph of the function $\displaystyle y = \sin \left({\frac 1 x}\right)$ for $x > 0$.

Let $J$ be the interval joining the points $\left({0, -1}\right)$ and $\left({0, 1}\right)$ in $\R^2$.

Then:

• $G$, is connected, but not path-connected.
• $G \cup J$ is connected.

Proof

• First we need to show that $G \cup J$ is connected.

Since the open interval $\left({0 .. \infty}\right)$ is connected, then so is $G$ by Continuous Image of Connected Space is Connected.

It is enough, from Subset of Closure of Connected Subspace, to show that $J \subseteq \operatorname{cl}\left({G}\right)$.

Let $p \in J$, say, $\left({0, y}\right)$ where $-1 \le y \le 1$.

We need to show that $\forall \epsilon > 0: N_\epsilon \left({p}\right) \cap G \ne \varnothing$, where $N_\epsilon \left({p}\right)$ is the $\epsilon$-neighborhood of $p$.

Let us choose $\displaystyle n \in \N: \frac 1 {2 n \pi} < \epsilon$.

From Sine of Multiple of Pi Plus Half:

$\displaystyle \sin \left({\frac {\left({4n + 1}\right) \pi} 2}\right) = 1$

and:

$\displaystyle \sin \left({\frac {\left({4n + 3}\right) \pi} 2}\right) = -1$

So by the Intermediate Value Theorem, $\displaystyle \sin \left({\frac 1 x}\right)$ takes every value between $-1$ and $1$ in the closed interval $\displaystyle \left[{\frac 2 {\left({4n + 3}\right) \pi} .. \frac 2 {\left({4n + 1}\right) \pi}}\right]$.

In particular, $\displaystyle \sin \left({\frac 1 {x_0}}\right) = y$ for some $x_0$ in this interval.

The distance between the points $\left({0, y}\right)$ and $\displaystyle \left({x_0, \sin \left({\frac 1 {x_0}}\right)}\right) = \left({x_0, y}\right)$ is $x_0 < \epsilon$.

So $\displaystyle \left({x_0, \sin \left({\frac 1 {x_0}}\right)}\right) \in N_\epsilon \left({p}\right) \cap G$, as required.

• Now we show that $G$ is not path-connected.

Let $I \subseteq \R$ be the closed interval $\left[{0 .. 1}\right]$.

Let $\displaystyle A = \left({\frac 1 \pi, 0}\right) \in \R^2$.

This proof is based on the fact that a continuous path $f: I \to G \cup J$ beginning at $A$ will never actually arrive at $0 = \left({0, 0}\right) \in \R^2$ because $I$ is compact.

Suppose $f: I \to G \cup J$ is continuous and that $f \left({0}\right) = A$.

Let $f_1: I \to \R$ and $f_2: I \to \R$ denote $\operatorname{pr}_1 \circ i \circ f$ and $\operatorname{pr}_2 \circ i \circ f$, where $i$ is the inclusion mapping and $\operatorname{pr}_1, \operatorname{pr}_2$ the first and second projections on the $x$ and $y$ axes.

Thus $f_2$ describes the vertical movement of the graph, and $f_1$ the horizontal movement.

Now $f_2$ is continuous and $I$ compact.

So by the Heine-Cantor Theorem, $f_2$ is uniformly continuous on $I$.

Let $\delta > 0$ be such that $\left|{f_2 \left({t}\right) - f_2 \left({t'}\right)}\right| < 2$ for any $t, t' \in I$ such that $\left|{t - t'}\right| < \delta$.

Let $0 < t_0 < t_1 < \ldots < t_n = 1$ be such that $t_i - t_{i-1} < \delta$ for all $i = 1, 2, \ldots, n$.

Now: as $t$ goes from $t_0$ to $t_1$, $f \left({t}\right)$ (which starts at $A$), can not reach a point $C$ where $y = 1$ without passing through a point $B$ where $y = -1$ on the way.

But then $f \left({t}\right) = B$ and $f \left({t'}\right) = C$ for some $t, t'$ where $\left|{t - t'}\right| < \delta$.

And since $B$ and $C$ are $2$ apart, $\left|{f_2 \left({t}\right) - f_2 \left({t'}\right)}\right| = 2$ which contradicts the choice of $\delta$.

Similarly, when going from $t_1$ to $t_2$, $t$ can similarly not get past more than one hump.

So, as $t$ goes from $0$ to $1$, $f \left({t}\right)$ can not traverse more than $n$ humps.

We formalize this discussion by induction.

We will show that:

$\displaystyle \forall i = 0, 1, \ldots, n: f_1 \left({t_i}\right) > \frac 2 {\left({2i + 3}\right) \pi}$

Since $\displaystyle f \left({t_0}\right) = f \left({0}\right) = \left({\frac 1 \pi, 0}\right)$, we have $\displaystyle f_1 \left({t_0}\right) = \frac 1 \pi > \frac {2} {3 \pi}$.

Suppose that $\displaystyle f_1 \left({t_i}\right) > \frac {2} {\left({2i + 3}\right) \pi}$ for some $i \ge 0$.

Suppose also, to get a contradiction, that $\displaystyle f_1 \left({t_{i+1}}\right) \ge \frac 2 {\left({2i + 5}\right) \pi}$.

Then since $f_1$ is continuous, by the I.V.P. there exists $t, t' \in \left[{t_1 .. t_{i+1}}\right]$ such that:

$\displaystyle f_1 \left({t_i}\right) = \frac {2} {\left({2i + 3}\right) \pi}, f_1 \left({t_{i+1}}\right) = \frac {2} {\left({2i + 5}\right) \pi}$

But the only point in $G \cup J$ whose first coordinate is $\displaystyle \frac {2} {\left({2i + 3}\right) \pi}$ is $\displaystyle \left({\frac 2 {\left({2i + 3}\right) \pi}, \sin \left({\frac {\left({2i + 3}\right) \pi} 2}\right)}\right)$.

So:

$\displaystyle f_2 \left({t}\right) = \sin \left({\frac {\left({2i + 3}\right) \pi} 2}\right)$

Similarly:

$\displaystyle f_2 \left({t'}\right) = \sin \left({\frac {\left({2i + 5}\right) \pi} 2}\right)$

Hence $\left|{f_2 \left({t}\right) - f_2 \left({t'}\right)}\right| = 2$ while $\left|{t - t'}\right| < \delta$.

This contradicts the choice of $\delta$.

This proves the induction.

The result follows from the fact that:

$\displaystyle f_1 \left({1}\right) > \frac 2 {\left({2n + 3}\right) \pi} > 0$, so $f_1 \left({1}\right) \ne 0$

$\blacksquare$