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Calculus and Beyond Homework Help
Proving outer content zero for a subset of the unit square
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[QUOTE="STEMucator, post: 4493598, member: 426227"] [h2]Homework Statement [/h2] 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. [h2]Homework Equations[/h2] ##C(S) = inf\{ \sum A_i \} = inf\{Area(P)\}## [h2]The Attempt at a Solution[/h2] 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##. [/QUOTE]
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Proving outer content zero for a subset of the unit square
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