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

filter54321

## Homework Statement

What is the closure of F = { (x,sin(1/x) : 0<x<=1 }?

None

## 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.

## Answers and Replies

Mathnerdmo
What do you know about sin(1/x)?

Look at the graph again. It does something very interesting.

filter54321
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
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
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.

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.