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I always liked the Sierpinski triangle, and happened upon the related article about the Sierpinski carpet.

The article is pretty sparse, and states the area of the carpet is zero (in standard Lebesgue measure).

Is there a proper proof or book in which a proof may be found?

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Lebesgue measure can be caluclated by subtracting measures of intevals $(1/3, 2/3)$ , $(1/9, 2/9)$ , $(7/9, 8/9)$ etc from measure of interval $(0,1)$. So if you want to calculate measure of carpet you should subtract measures of $(1/3,2/3)\times(1/3,2/3)$, $(1/9, 2/9)\times(1/9,2/9)$, $(1/9, 2/9)\times(7/9,8/9)$, $(7/9, 8/9)\times(1/9,2/9)$, $(7/9, 8/9)\times(7/9,8/9)$ etc from measure of $(0,1)\times(0,1)$ – Norbert Nov 19 '11 at 10:07
Think about how much area you're removing at each stage of the carpet's construction, and see what all those areas summed up converge to. (In short, prove this: the area that remains after $n$ steps of the carpet's construction is $\left(\dfrac89\right)^n$. What does this converge to as $n\to \infty$?) – J. M. Nov 19 '11 at 10:09
You should look into Hausdorff measure and dimension. The Hausdorff dimension of the carpet is higher than 1 and less than 2. So the area is 0 and the length is infinite. – Joe Moeller Jun 12 '15 at 7:25

If after $n$ iterations for the square the area of the carpet is $A_n$, then you have the recursion $A_{n+1}=\dfrac{8}{9} A_n$, so $A_{n}=\left(\frac{8}{9}\right)^n A_0$ and the limit of the area is $0$ as $n$ increases.

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Oh, that's pretty simple. Thanks. – Muencher Nov 19 '11 at 10:35
@Muencher, if you feel that Henry's answer is satisfactory, please click on the check mark on the left of Henry's answer. – J. M. Nov 20 '11 at 10:05

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