What can we say about the graph of $\sin \frac 1x$ close to $0$?

Question

Let $\varepsilon > 0$ small. Let $f(x) = \sin \frac 1x$ with $f(0) = 0$. Consider, as a subset of $\mathbb{R}^2$, the graph of $f(x)$ in the interval $(-\varepsilon, \varepsilon)$; let us denote this by $G_{\varepsilon}$. Now, consider

$$S = \left\{(x,y) : x \in (-\varepsilon, \varepsilon), y \in [-1, 1] \right\}.$$

Clearly, for all $\varepsilon > 0$, $G_{\varepsilon}$ is not dense in $S$. However, nevertheless, there is a strong intuitive feeling (based on the visuals of the graph) that as $\varepsilon \to 0$, $S \setminus G_{\varepsilon}$ gets "sparser".

Is there a way (perhaps topological) to formally describe this intuition?

Byt the fact that $\overline{G_\epsilon}=G_\epsilon\cup({0}\times[-1,1])$? — Did, Oct 17 '16 at 05:53
The "distance" between each new wave (which is what you intuitively interpret as "getting denser") is not a topological notion. So if you want to formally describe it, you would at least have to go to real analysis, where distance is a thing. However, note that the graph is infinitely thin, so for instance the area of $S\setminus G_\epsilon$ is the same all the way, namely the area of $S$. So in that sense it's not getting sparser at all. Maybe the area of the "biggest slice" could capture what you're after. — Arthur, Oct 17 '16 at 06:38

score 0 · Answer 1 · answered Oct 17 '16 at 18:16

Suppose you draw the curve $y=\sin(\frac1{x})$ with a calligraphic pen of positive width $w,$ and you don't want the entire rectangle $[-h,h]\times[-1,1]$ to be filled in (where $h$ is some small positive number). Then the width $w$ of the pen has to be tiny, less than $2\pi h^2.$

Notice that if you reduce the width of the rectangle that you don't want to be entirely filled in, the width of the pen has to be reduced by a much larger amount proportionally.

Here's why this is true: $\sin x$ is periodic with period $2 \pi,$ so there will be both a peak and a trough for $\sin x$ somewhere in the half-open interval $[\dfrac{1}{h},\dfrac{1}{h}+2\pi).$ It follows that there will be both a peak and a trough for $\sin(1/x)$ somewhere in the half-open interval $(\dfrac1{\frac{1}{h}+2\pi},h].$ The width of this latter half-open interval is

\begin{align} h-\frac1{\frac{1}{h}+2\pi}&=\frac{2\pi h}{\frac{1}{h}+2\pi} \\&\le \dfrac{2\pi h}{1/h} \\&= 2\pi h^2, \end{align}

so a pen that wide will cover the entire vertical line $\lbrace \frac1{n} \rbrace \times [-1,1].$

Now, $\sin (1/x)$ gets even more compressed as you get closer to the origin, so the same width pen will cover $(0,\frac1{n}]\times[-1,1].$ By symmetry, the same pen will also cover $[-\frac1{n},0)\times [-1,1].$ Finally, the pen covers $\lbrace 0 \rbrace \times [-1,1],$ because the curve has both peaks and troughs with $x$ closer than the pen width to the origin.

What can we say about the graph of $\sin \frac 1x$ close to $0$?

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