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

by filter54321
Tags: 0<x<1, closure, sin1 or x
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filter54321
#1
Mar1-10, 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
Sclosure = dS U S

Therefore Fclosure = 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.
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Mathnerdmo
#2
Mar1-10, 06:34 PM
P: 41
What do you know about sin(1/x)?

Look at the graph again. It does something very interesting.
filter54321
#3
Mar1-10, 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?

JG89
#4
Mar1-10, 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.
Mathnerdmo
#5
Mar1-10, 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.

Quote Quote by filter54321 View Post
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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