引理敘述
有限多球
![維塔利覆蓋引理](/img/2/729/wZwpmLyADN3gTM5ETMxcTN1UTM1QDN5MjM5ADMwAjMwUzLxEzL2EzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/7/423/wZwpmL4YTO0cTMwADMxcTN1UTM1QDN5MjM5ADMwAjMwUzLwAzL4UzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
在一個度量空間中有一族閉球,則這一族球中存在互不相交的球,適合條件
![維塔利覆蓋引理](/img/8/737/wZwpmLwUDNwEjNwMDOwcTN1UTM1QDN5MjM5ADMwAjMwUzLzgzLzczLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/3/c58/wZwpmL3QDNxgDO5MTOwcTN1UTM1QDN5MjM5ADMwAjMwUzLzkzLyMzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/b/a05/wZwpmL0UDNwQDM2cTOwcTN1UTM1QDN5MjM5ADMwAjMwUzL3kzLyUzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/b/a05/wZwpmL0UDNwQDM2cTOwcTN1UTM1QDN5MjM5ADMwAjMwUzL3kzLyUzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
表示和有相同中心,而半徑是的三倍的球。
無限多球
![維塔利覆蓋引理](/img/a/35a/wZwpmL3cDN4cTO0ADMxcTN1UTM1QDN5MjM5ADMwAjMwUzLwAzL4IzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
在一個度量空間中有一族半徑為正數的閉球,這族球的半徑有有限的上界,即
![維塔利覆蓋引理](/img/8/5be/wZwpmL1QjN2gDN0cTOwcTN1UTM1QDN5MjM5ADMwAjMwUzL3kzLwUzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/d/68c/wZwpmLxcDO2cDN1IDOwcTN1UTM1QDN5MjM5ADMwAjMwUzLygzLyAzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/f/64e/wZwpmLzQTNzETM1UDOwcTN1UTM1QDN5MjM5ADMwAjMwUzL1gzL2QzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
則這一族球中存在互不相交的球,,適合條件
![維塔利覆蓋引理](/img/8/4aa/wZwpmLxAjMwgDO4IDOwcTN1UTM1QDN5MjM5ADMwAjMwUzLygzL0QzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/4/ad6/wZwpmLzcjN3UDN4IDMxcTN1UTM1QDN5MjM5ADMwAjMwUzLyAzLzgzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/a/3b7/wZwpmLyIjNwMzN1kzN0YzM1UTM1QDN5MjM5ADMwAjMwUzL5czLxUzLt92YucmbvRWdo5Cd0FmLwE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/a/3b7/wZwpmLyIjNwMzN1kzN0YzM1UTM1QDN5MjM5ADMwAjMwUzL5czLxUzLt92YucmbvRWdo5Cd0FmLwE2LvoDc0RHa.jpg)
表示和有相同中心,而半徑是的五倍的球。
證明
有限情形
![維塔利覆蓋引理](/img/7/cba/wZwpmL1MDOwQzM2kTOwcTN1UTM1QDN5MjM5ADMwAjMwUzL5kzL4UzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/7/cba/wZwpmL1MDOwQzM2kTOwcTN1UTM1QDN5MjM5ADMwAjMwUzL5kzL4UzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/b/fd9/wZwpmLwgjNzQzM4cDMxcTN1UTM1QDN5MjM5ADMwAjMwUzL3AzLzYzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/b/a05/wZwpmL0UDNwQDM2cTOwcTN1UTM1QDN5MjM5ADMwAjMwUzL3kzLyUzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/b/a05/wZwpmL0UDNwQDM2cTOwcTN1UTM1QDN5MjM5ADMwAjMwUzL3kzLyUzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/3/c58/wZwpmL3QDNxgDO5MTOwcTN1UTM1QDN5MjM5ADMwAjMwUzLzkzLyMzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
取這一族球中半徑最大的一個球,然後除去所有與相交的球。再從剩下的球中取半徑最大的為,如此類推。那么任何其他的球必定因為和某個相交而被除去,這個球的半徑不大於,因此包含在之內。
無限情形
![維塔利覆蓋引理](/img/b/01d/wZwpmLyIjNwATN5MTOwcTN1UTM1QDN5MjM5ADMwAjMwUzLzkzL2YzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/b/01d/wZwpmLyIjNwATN5MTOwcTN1UTM1QDN5MjM5ADMwAjMwUzLzkzL2YzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/e/5b1/wZwpmL3EjN5kDM3MTMxcTN1UTM1QDN5MjM5ADMwAjMwUzLzEzLxgzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/b/17e/wZwpmL4YDO2ETMycTOwcTN1UTM1QDN5MjM5ADMwAjMwUzL3kzL1IzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
設這一族球的半徑的上確界為 R。將這一族按半徑分成子集, j為正整數;包含半徑在區間的球。依次取如下:
![維塔利覆蓋引理](/img/6/480/wZwpmL2AzN4ADNygDMxcTN1UTM1QDN5MjM5ADMwAjMwUzL4AzL3QzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/b/72b/wZwpmLwcDMzUTO3IDOwcTN1UTM1QDN5MjM5ADMwAjMwUzLygzL0YzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/a/db8/wZwpmL0gzN0kzM2kTOwcTN1UTM1QDN5MjM5ADMwAjMwUzL5kzL3UzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/a/db8/wZwpmL0gzN0kzM2kTOwcTN1UTM1QDN5MjM5ADMwAjMwUzL5kzL3UzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/b/72b/wZwpmLwcDMzUTO3IDOwcTN1UTM1QDN5MjM5ADMwAjMwUzLygzL0YzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
設。取為內互不相交球的子集之中的極大者,即其他在中的球都與這一子集中某個球相交。從佐恩引理知這樣的存在,以下同。
![維塔利覆蓋引理](/img/3/e2c/wZwpmLwgDN0kzN1YDMxcTN1UTM1QDN5MjM5ADMwAjMwUzL2AzL0gzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/f/43e/wZwpmL2IzNykjMwMTOwcTN1UTM1QDN5MjM5ADMwAjMwUzLzkzL4UzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/b/c24/wZwpmLxAzM4cTOxETMxcTN1UTM1QDN5MjM5ADMwAjMwUzLxEzL4IzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/8/ff9/wZwpmLxUzM2YjMwUDOwcTN1UTM1QDN5MjM5ADMwAjMwUzL1gzL2AzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/8/1f5/wZwpmL2QTOxEDMwcDMxcTN1UTM1QDN5MjM5ADMwAjMwUzL3AzL3AzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/f/43e/wZwpmL2IzNykjMwMTOwcTN1UTM1QDN5MjM5ADMwAjMwUzLzkzL4UzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
設已取, k為某大於1的整數。設是中不與中任何球相交的全部球的子集。取為內互不相交球的子集之中的極大者。
![維塔利覆蓋引理](/img/1/09e/wZwpmLwYTO4MjM5QDOwcTN1UTM1QDN5MjM5ADMwAjMwUzL0gzL3MzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/f/43e/wZwpmL2IzNykjMwMTOwcTN1UTM1QDN5MjM5ADMwAjMwUzLzkzL4UzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/8/1f5/wZwpmL2QTOxEDMwcDMxcTN1UTM1QDN5MjM5ADMwAjMwUzL3AzL3AzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/e/526/wZwpmL0IjNzUTM0MDOxUTN1UTM1QDN5MjM5ADMwAjMwUzLzgzLzMzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/e/526/wZwpmL0IjNzUTM0MDOxUTN1UTM1QDN5MjM5ADMwAjMwUzLzgzLzMzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/2/227/wZwpmL0MDNxADO2kDMxcTN1UTM1QDN5MjM5ADMwAjMwUzL5AzLxIzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
設。任何其他的球 B必在某一個中,因此這個球與中一個球相交,而的半徑大於 B的半徑的二分之一,故此B包含在之內。
討論
![維塔利覆蓋引理](/img/b/01d/wZwpmLyIjNwATN5MTOwcTN1UTM1QDN5MjM5ADMwAjMwUzLzkzL2YzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/7/0f6/wZwpmLzAjMzMDM1ETMxcTN1UTM1QDN5MjM5ADMwAjMwUzLxEzL4gzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
因為有無限多球時,可能不存在半徑最大的球,所以在構造中,每一步選擇的球的半徑,只要求接近餘下的球的半徑的上確界。而結果中的5並非最佳常數。將的定義中的的2換成任何大於1的數c,那么就可把結果中的5換成1+2c,即可以用任何大於3的數取代。不過由於未必有半徑最大的球,以致不能像有限多球時用3取代,以下是一個簡單例子。
![維塔利覆蓋引理](/img/1/52e/wZwpmLwADNxUDO5kDM5YzM1UTM1QDN5MjM5ADMwAjMwUzL5AzL4czLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/9/463/wZwpmL1EjN2kjM1YjMxMzM1UTM1QDN5MjM5ADMwAjMwUzL2IzLzIzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/f/5e9/wZwpmL1ADM1QTN3UDOwcTN1UTM1QDN5MjM5ADMwAjMwUzL1gzL0EzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/9/463/wZwpmL1EjN2kjM1YjMxMzM1UTM1QDN5MjM5ADMwAjMwUzL2IzLzIzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/d/793/wZwpmLxYTN4ITM3kTOwcTN1UTM1QDN5MjM5ADMwAjMwUzL5kzLwgzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/c/f97/wZwpmLyITM5MDO1MDMxcTN1UTM1QDN5MjM5ADMwAjMwUzLzAzLwYzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/9/463/wZwpmL1EjN2kjM1YjMxMzM1UTM1QDN5MjM5ADMwAjMwUzL2IzLzIzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/7/1dc/wZwpmLzUDN4kDO2czNxMzM1UTM1QDN5MjM5ADMwAjMwUzL3czL0IzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/a/7b1/wZwpmL3YjMyMDN3ADMxcTN1UTM1QDN5MjM5ADMwAjMwUzLwAzL2QzLt92YucmbvRWdo5Cd0FmLwE2LvoDc0RHa.jpg)
![維塔利覆蓋引理](/img/7/1dc/wZwpmLzUDN4kDO2czNxMzM1UTM1QDN5MjM5ADMwAjMwUzL3czL0IzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
在平面中,給出如下的一族球:對每個正整數n,是半徑為的閉球,若n為奇數,的圓心在;若 n為偶數,則圓心在。所有球都包含原點(0,0),故任意兩個球都相交,因此包含互不相交的球的子集只能有一個球。這一族球的半徑上確界是2,然而全部球的半徑都小於2。若選任何一個為這個子集,因有半徑更大的球在原點的另一側,故此不覆蓋。
套用
這條引理可用於證明哈代-李特爾伍德極大不等式。
參見
•貝西科維奇覆蓋定理