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

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In summary: For example, take the point (0,0). Every open ball around this point will contain points both in F and in the complement of F, so (0,0) is in the boundary. This means that the closure of F, which is the union of F and its boundary, will include (0,0) in addition to the points in F. Therefore, the closure of F is not equal to F.In summary, the closure of F = {(x, sin(1/x) : 0 < x ≤ 1} is not equal to F because there are points in the boundary of F that are not in F itself. This is due to the oscillating behavior of sin(1/x)
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filter54321
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Homework Statement


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


Homework Equations


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.
 
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  • #2
What do you know about sin(1/x)?

Look at the graph again. It does something very interesting.
 
  • #3
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
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
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.

filter54321 said:
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.
 

1. What is the closure of { (x,sin(1/x) : 0

The closure of a set is the smallest closed set that contains all the points in the original set. In this case, the closure of { (x,sin(1/x) : 0

2. How is the closure of { (x,sin(1/x) : 0

The original set { (x,sin(1/x) : 0

3. How is the closure of { (x,sin(1/x) : 0

The closure of a set is closely related to the concept of continuity. A function is continuous at a point if and only if the limit of the function at that point is equal to the value of the function at that point. In this case, the closure of { (x,sin(1/x) : 0

4. How can the closure of { (x,sin(1/x) : 0

The closure of { (x,sin(1/x) : 0

5. What is the significance of the closure of { (x,sin(1/x) : 0

The closure of a set is a fundamental concept in topology and analysis. It allows us to study the properties of a set and its limits, and is essential in the study of continuity, convergence, and compactness. In this specific case, the closure of { (x,sin(1/x) : 0

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