# Closure of { (x,sin(1/x) : 0<x<=1 }?

by filter54321
Tags: 0<x<1, closure, sin1 or x
 P: 39 1. The problem statement, all variables and given/known data What is the closure of F = { (x,sin(1/x) : 0
 P: 41 What do you know about sin(1/x)? Look at the graph again. It does something very interesting.
 P: 39 I know what it looks like, it's a common example in calculus. It bounces up and down as you go to 0 with ever increasing frequency. How does that have anything to do with the boundary methodology outlined in my original post?
P: 720

## Closure of { (x,sin(1/x) : 0<x<=1 }?

Use the fact that Cl(F) = lim(F) = {x : x is a limit of F}. Now let x approach any value in (0,1] and look at the set of F's limits.
P: 41
The graph of y = sin(1/x), in particular that it oscillates with increasing frequency as x gets closer to 0, has everything to do with the boundary.

 Quote by filter54321 F is a squiggly line in R2. For every point in F (every point on the squiggly line) an open ball about that point will contain point both in F and in the complement of F. Therefore F is it's own boundary.
The problem is right here. All you've shown is that F is a subset of its boundary. There may be other points of R2 in the boundary.

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