closure of the set $S=\{(x,y)|x\in(0,1)\cap\mathbb{Q};y=\sin(\frac1x)\}$

Question

I am trying to determine the closure of the set $S=\{(x,y)|x\in(0,1)\cap\mathbb{Q};y=\sin(\frac1x)\}$.

I don't know where to start. If we define a sequence $(x_n,y_n)=(x_n,\sin(\frac{1}{x_n})),x\in(0,1)\cap\mathbb{Q}$ then I think it converges for all $(x_n)$ except those which converges to $0$ because $\sin(\frac{1}{x_n})$ is not cauchy for $(x_n)\to0$. But $0\not\in S.$ So, can I take the set of all limit points as the whole set $S$?

Answer 1

The good thing about sequences in $\mathbb{R^2}$ is that they are convergent if and only if they converge in each coordinate. It is obvious that if $(x_n,y_n)\to (x,y)$ when $\{(x_n,y_n)\}$ is a sequence of vectors in $S$ then $x\in [0,1]$. The question is what is $y$. Well, if $x\in (0,1]$ then obviously by continuity of sine we must have $y=\sin(\frac{1}{x})$. Also, it is known that for each $x\in (0,1]$ you can find a sequence of rational numbers from $(0,1)$ that converge to $x$. Hence any point of the type $(x,\sin\frac{1}{x})$ when $x\in (0,1]$ is in the closure of $S$.

Now, I say any point of the form $(0,y)$ when $y\in [-1,1]$ is in the closure of $S$ as well. I suppose you know how the function $f(x)=\sin\frac{1}{x}$ acts in the neighborhood of $0$. For each $n\in\mathbb{N}$ we have $f([\frac{1}{2\pi(n+1)},\frac{1}{2\pi n}])=[-1,1]$. So for each $y\in [-1,1]$ you can find a sequence of positive numbers $x_n$ such that $f(x_n)=y$ for all $n$. Then the limit of $(x_n,f(x_n))$ will be $(0,y)$. As $(0,y)$ is a limit of a sequence of elements in the closure of $S$ then we conclude it is also in the closure of $S$.

Answer 2

Sometimes the best strategy is to proceed in steps.

The first step would be to notice that the closure of $S$ contains the entire set $T = \{(x,y) \mid x \in (0,1]; y = \sin(\frac{1}{x})\}$, not just for $x \in (0,1) \cap \mathbb Q$ but for all $x \in (0,1]$. The reason for this is twofold: $(0,1) \cap \mathbb Q$ is dense in $(0,1]$ so each $x \in (0,1]$ is a limit of a sequence $(x_n)$ in $(0,1) \cap \mathbb Q$; and the function $y=\sin(\frac{1}{x})$ is continuous so $\sin(\frac{1}{x})$ is the limit of $\sin(\frac{1}{x_n})$.

The next step requires some visual imagination. The set $T$ is literally the graph of the function $y=\sin(\frac{1}{x})$ over the interval $0 < x \le 1$. If you attempt to draw that graph, you will see that it squiggles up and down more and more tightly, approaching the vertical interval $\{0\} \times [-1,+1]$. This should lead you to a mathematical conjecture: the interval $\{0\} \times [-1,+1]$ is contained in the closure of the set $T$, and therefore is also contained in the closure of the set $S$. Now try to prove that this conjecture is true, and once you've done that then you will know that the set $U = T \cup (\{0\} \times [-1,+1])$ is contained in the closure of $S$.

The final step is to prove that $U$ is closed, at which point you will know that $U$ is equal to the closure of $S$.