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Math Help - Justifying the closure of a set

  1. #1
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    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.
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  2. #2
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    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.
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  3. #3
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    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.
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  4. #4
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    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.
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