Proving outer content zero for a subset of the unit square

[STEMucator]

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 $C(S) = 0$.

 

[Dick]

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.

 

[STEMucator]

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.