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Closure of { (x,sin(1/x) : 0<x<=1 }? 
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#1
Mar110, 06:28 PM

P: 39

1. The problem statement, all variables and given/known data
What is the closure of F = { (x,sin(1/x) : 0<x<=1 }? 2. Relevant equations None 3. The attempt at a solution 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 closure of a set is equal to the unions of the boundary of the set and the set itself S_{closure} = dS U S Therefore F_{closure} = dF U F = F U F = F My professor indicated that this line of reasoning is flawed. I'm not sure why nor am I sure what a correct anyswer would be. Any help would be appreciated. 


#2
Mar110, 06:34 PM

P: 41

What do you know about sin(1/x)?
Look at the graph again. It does something very interesting. 


#3
Mar110, 07:08 PM

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?



#4
Mar110, 07:08 PM

P: 726

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.



#5
Mar110, 08:10 PM

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.



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