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集合S和S′之間的對應用以下的方式來建立。命x_0為集合S的某一點,這點屬於某個一級矩形;又属於某個二级矩形,(這二級矩形包在前面的一級矩形之內);又屬於某個三級矩形,(這三級矩形父包在前面的二級矩形之內),如此這般等樣,叫這些矩形中每個矩形與謝爾平斯基鋪蓋構造中的正方形相對應,這些正方形與我們所考慮的矩形同級,而且它在謝爾平斯基鋪蓋S′的基本正方形Q′_0中所處的位置與該矩形在Q_0中所處的位置一樣。於是我們便得一系列正方形,其中每個後面的正方形都包地前面的正方形內,而且這些正方形的邊長將隨共級數之增加而趨於零,因此所有這些正方形
The pair between the sets S and S’ is established in the following manner. The life x_0 is a certain point of the set S, which belongs to a certain first-order rectangle; it also belongs to a certain second-level rectangle, (this second-level rectangular packet is within the front-level rectangle); it also belongs to a three-level rectangle, (this The three-level rectangular parent is wrapped in the previous two-level rectangle), so that each rectangle in these rectangles corresponds to the square in the Shelpinski blanket structure, which is at the same level as the rectangle we are considering. And it is located in the basic square Q′_0 of the Schepinski blanket S′ as it is in Q_0. So we have a series of squares, each of which is surrounded by squares in front of the square, and the length of the sides of these squares will tend to zero as the number of co-levels increases, so all these squares