# Thread: Justifying the closure of a set

1. ## Justifying the closure of a set

My problem is the following:
Find $\overline{A} \textrm{ for } A=\{(x,\sin(1/x))|x>0.\}$
I know that $\overline{A}=A\cup\{(0,y)|-1\leq y \leq 1,\}$ 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 $\{0\}\times [-1,1]$ is the set of limit points of $A$.
For any point $(0,t),~-1\le t\le 1$ and any $\rho >0$
does the interior of the disk $x^2+(y-t)^2<\rho^2$ contain a point of $A?$
If so, $(0,t)$ 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 $-1\le y\le 1$, 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, $sin(z+ 2n\pi)= sin(z)= y$ so that letting $x_n= \frac{1}{z+ 2n\pi}$, $x_n$ goes to 0 as n goes to infinity and $sin(\frac{1}{x})= sin(z+ 2n\pi)$ goes to y: the points $(x_n, sin(\frac{1}{x_n})$ converge to (0, y) so (0, y) is a limit point.