# $\sin(\frac{1}{x})$ on $\Bbb R^2$ interior, cluster points, boundary

Given $$M:=\{(x_1, x_2) \in \Bbb R^2 : x_2 =\sin(\frac 1 {x_1}), x_1\in(0,\frac 1 \pi)\}$$ , a subset of the normed Space $(\Bbb R^2,||\cdot||_2)$.

$M$ is the above Curve. (the commen $\sin(\frac{1}{x})$ function)

My Problem is the following: I have to determine the interior, the cluster points and the boundary of $M$.

If $M$ were a normal curve I would say that the Interior is the empty set, every point of $M$ is a cluster point (in addition to the points of $x_1 = 0$ and $\frac{1}{\pi}$) and that the boundary of $M$ is $M$ (again in addition to the above points). Now since $\sin(\frac{1}{x})$ would be infinity for $x_1=0$ I excluded this point from the boundary and the cluster points.

The only thing that's left to consider now is the infinitly fast oszilation of the curve aproaching $x_1=0$. What effect does this oszillation have regarding the interior, the cluster points and the boundary? Also: is $M$ bounded?

• My thoughts were, since it is still a curve if you look close enough, the oszilation shouldn't pose a problem, but the singularity for x1=0 does, but since that point isn't included in M maybe it doesn't. Any thoughts? – DeltaChief Apr 22 '16 at 19:02

A point $p$ of $M$ is an interior point if there exists some open ball $B_{\epsilon}$ centred at $p$ which is completely contained in $M$. Bearing this in mind, if take any point $p$ of $M$, then it's fairly straightforward to show that there are points arbitrarily close to $p$ not contained in $M$. Hence, $M$ has no interior points and, as you say, the interior is the empty set.
A cluster point of $M$ is one that can be approximated by a sequence of points in $M$. Equivalently, a point $p$ is a cluster point of $M$ if for any $\epsilon >0$ there is a ball of radius $\epsilon$ centred at $p$ which intersects $M$. Going with this last definition, as you say, any point of $M$ is a cluster point. Also, $\left(\frac{1}{\pi}, 0 \right)$ is a limit point on the right. Perhaps most interesting are the points on the left. Any point which satisfies $x_1=0, |x_2| \le 1$ can also be obtained as the limit of some sequence in $M$ so these points are also cluster points. So, in summary, the cluster points are $$M \,\,\cup \,\, \left(\frac{1}{\pi}, 0 \right) \,\, \cup \,\, \{(0,x_2) \in \mathbb{R}^2 : |x_2| \le 1 \}$$
$M$ is clearly bounded as you can tell from your plot.