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.
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?
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.
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. 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.