# Thread: Justifying the closure of a set

1. ## Justifying the closure of a set

My problem is the following:
Find $\displaystyle \overline{A} \textrm{ for } A=\{(x,\sin(1/x))|x>0.\}$
I know that $\displaystyle \overline{A}=A\cup\{(0,y)|-1\leq y \leq 1,\}$ but I need to figure out & understand the how'' behind that answer, in other words, give a proof that A-closure is that set above. Thanks.

2. Do you understand that the closure of a set is the union of the set with all of its limit points?
If so then all we have to do is show that $\displaystyle \{0\}\times [-1,1]$ is the set of limit points of $\displaystyle A$.
For any point $\displaystyle (0,t),~-1\le t\le 1$ and any $\displaystyle \rho >0$
does the interior of the disk $\displaystyle x^2+(y-t)^2<\rho^2$ contain a point of $\displaystyle A?$
If so, $\displaystyle (0,t)$ is a limit point.

3. I partially understood what you just said. However, some of the things I am having trouble with are the notions of closure & limit points. I do understand the first statement, though. But some of the rest I am a bit vague on. Sorry, but I really am trying to understand these ideas as much as possible.

4. Have you graphed the function y= sin(1/x)? Close to x= 0, the graph oscillates infinitely many times between -1 and 1. For any point (0, y) with $\displaystyle -1\le y\le 1$, there exist points on the graph arbitrarily close to (0, y). That's why any such point is in the closure of the graph.

In particular, given y between -1 and 1, there exist some z such that sin(z)= y. Then also, $\displaystyle sin(z+ 2n\pi)= sin(z)= y$ so that letting $\displaystyle x_n= \frac{1}{z+ 2n\pi}$, $\displaystyle x_n$ goes to 0 as n goes to infinity and $\displaystyle sin(\frac{1}{x})= sin(z+ 2n\pi)$ goes to y: the points $\displaystyle (x_n, sin(\frac{1}{x_n})$ converge to (0, y) so (0, y) is a limit point.