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I have a set $$A=\{(x,y)\in\mathbb{R^2}\mid x>0,y=\sin\left(\frac{1}{x}\right)\}$$ and I want to find a closure of this set. In the first place I thought, that the closure ($Cl(A)$) is equal to $X=A\cup\{(0,0)\}$, but I strongly doubt about this. I am absolutely sure that $X\subset Cl(A)$.

Could you give me any hints that will help me to come to the answer?

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Have you looked at the graph of $y=\sin(1/x)$? In particular, note it "approaches" the segment $[-1,1]$ of the $y$-axis. – David Mitra Feb 16 '13 at 15:52
I know this is not your question, but note that this example is very well-known as it leads to the typical example of a connected space which is not arcwise connected. – 1015 Feb 16 '13 at 16:11
@David Mitra thx – Oiale Feb 16 '13 at 16:17
@julien Thanks for your link, but what does you mean when you said that it is not my question? I found this question in my textbook. – Oiale Feb 16 '13 at 16:19
Your question asks for closure and has nothing to do with connectedness. So my link does not help here, of course. But I thought you might wanted to know that this example comes around for other reasons. – 1015 Feb 16 '13 at 16:22
up vote 4 down vote accepted

Given $(x,y) \in A$ consider the sequence $\{(x_n,y_n)\}\subset A$ with $$ \frac{1}{x_n}=2n\pi +\frac{1}{x}. $$ Since $y_n=\sin(x^{-1})=y$ for every $n$, we have $(x_n,y_n) \to (0,y) \in \{0\}\times[-1,1]$. Hence $$ \{0\}\times[-1,1] \subset \text{Cl}(A). $$ Conversely if $(x,y) \in \text{Cl}(A)$, then there exists some sequence $\{(x_n,y_n)\} \subset A$ such that $(x_n,y_n) \to (x,y)$, in particular $x \ge 0$. If $x>0$ then $(x,y) \in A$, if not then $(x,y) \in \{0\}\times[-1,1]$. Thus $\text{Cl}(A)=A\cup\{0\}\times[-1,1]$.

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Thank you for so detailed answer. – Oiale Feb 16 '13 at 16:24

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