Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
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. planetmath.org/encyclopedia/TopologistsSineCurve.html –  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

1 Answer 1

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 $\text{Cl}(A)=A\cup\{0\}\times[-1,1]$.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.