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Unit Square & Outer Content

  • Thread starter Zondrina
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  • #1
Zondrina
Homework Helper
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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. Lets 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##.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618

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. Lets 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.
 
Last edited:
  • #3
Zondrina
Homework Helper
2,065
136
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.
 

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