Proving outer content zero for a subset of the unit square

In summary, the homework statement asserts that if a subset of the unit square has outer content zero, then the subset has that property. The Attempt at a Solution tries to solve this problem by outlining a method for determining whether or not a given subset has outer content zero. If the subset does have outer content zero, then the statement is proven. If the subset does not have outer content zero, then the statement is not proven.
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Homework Statement



Let ##S = \{ (x,y) \space | \space x=\frac{1}{n}, n = 1, 2, 3,... \}## be a subset of the unit square.

Prove ##S## has outer content zero.

Homework Equations



##C(S) = inf\{ \sum A_i \} = inf\{Area(P)\}##

The Attempt at a Solution



There are no answers I can check my work with so I don't know if this is okay. So I started drawing everything out ( really crowded drawing ) and I believe I know what I need to do.

##\forall ε > 0##, pick an ##N## such that ##\frac{1}{N} < \frac{ε}{2}##.

Then for ##n>N##, each line ##x = \frac{1}{n}## that I draw for ##n = 1, 2, 3...## will be inside a rectangle with coordinates ##[0, \frac{1}{N}] \times [0,1]## ( y only goes up to 1 because this is the unit square ).

For ##n ≤ N## and ##δ > 0##, choose an interval ##[\frac{1}{n} - δ, \frac{1}{n} + δ]## such that ##2Nδ < \frac{ε}{2} \Rightarrow δ < \frac{ε}{4N}##.

Now partition the unit square with a line at ##x = 0## ( Since ##n≠0## ) and then more lines at ##x = \frac{1}{n} ± δ## yielding a messy looking mesh by now. Let's call this partition ##P##. Then ##Area(P)## depends only on portions of the partition containing points from ##S##.

Therefore ##Area(P) = \frac{1}{N} + 2Nδ < \frac{ε}{2} + \frac{2Nε}{4N} = ε##.

Hence ##C(S) = inf\{Area(P)\} < ε, \forall ε > 0## so that ##C(S) = 0##.
 
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  • #2
Zondrina said:

Homework Statement



Let ##S = \{ (x,y) \space | \space x=\frac{1}{n}, n = 1, 2, 3,... \}## be a subset of the unit square.

Prove ##S## has outer content zero.

Homework Equations



##C(S) = inf\{ \sum A_i \} = inf\{Area(P)\}##

The Attempt at a Solution



There are no answers I can check my work with so I don't know if this is okay. So I started drawing everything out ( really crowded drawing ) and I believe I know what I need to do.

##\forall ε > 0##, pick an ##N## such that ##\frac{1}{N} < \frac{ε}{2}##.

Then for ##n>N##, each line ##x = \frac{1}{n}## that I draw for ##n = 1, 2, 3...## will be inside a rectangle with coordinates ##[0, \frac{1}{N}] \times [0,1]## ( y only goes up to 1 because this is the unit square ).

For ##n ≤ N## and ##δ > 0##, choose an interval ##[\frac{1}{n} - δ, \frac{1}{n} + δ]## such that ##2Nδ < \frac{ε}{2} \Rightarrow δ < \frac{ε}{4N}##.

Now partition the unit square with a line at ##x = 0## ( Since ##n≠0## ) and then more lines at ##x = \frac{1}{n} ± δ## yielding a messy looking mesh by now. Let's call this partition ##P##. Then ##Area(P)## depends only on portions of the partition containing points from ##S##.

Therefore ##Area(P) = \frac{1}{N} + 2Nδ < \frac{ε}{2} + \frac{2Nε}{4N} = ε##.

Hence ##C(S) = inf\{Area(P)\} < ε, \forall ε > 0## so that ##C(S) = 0##.

That's a little hard to read, but I think you've got the right idea. Draw a small rectangle including an infinite number of segments near x=0 and then bracket the remaining finite number of segments with very small rectangles. Seems pretty ok to me.
 
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  • #3
Dick said:
That's a little hard to read, but I think you've got the right idea. Draw a small rectangle including an infinite number of segments near x=0 and then bracket the remaining finite number of segments with very small rectangles. Seems pretty ok to me.

Thanks for the suggestion Dick, that makes it feel a little more rigorous. I was worried I would never see if I had the right idea or not.
 

1. What is a unit square?

A unit square is a square with sides that are each one unit in length. It has four equal sides and four right angles, making it a regular quadrilateral.

2. How is the area of a unit square calculated?

The area of a unit square is calculated by multiplying the length of one side by itself. In other words, the area is equal to the side length squared. Since the side length of a unit square is one, the area is simply one square unit.

3. What is the outer content of a unit square?

The outer content of a unit square is the perimeter, or the distance around the outside of the square. It is equal to the sum of all four sides.

4. How is the outer content of a unit square calculated?

The outer content of a unit square is calculated by adding together all four side lengths. Since each side is one unit in length, the outer content of a unit square is simply four units.

5. What is the relationship between the area and outer content of a unit square?

The area and outer content of a unit square are not directly related. However, the outer content can be thought of as the perimeter of a square with an area of one square unit. Additionally, the outer content can be used to calculate the length of the sides of a unit square by dividing it by four.

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