91影视

Let 鈩� be any element in S. Since n鈧� 鈭� 鈩� and S is a non-empty subset of T = {n 鈭� 鈩� | n 鈮� n鈧�}, we know that 鈩� 鈮� n鈧�. Thus, (鈩� - n鈧� + 1) is an element of T* = {n - n鈧� + 1 | n 鈭� T}. Let 鈩�* be the least element of T*. By definition, 鈩�* 鈮� (鈩� - n鈧� + 1) for all 鈩� in S. Now, we will show that n鈧� + 鈩�* - 1 is an element of S by proving that there exists an element 鈩�* in S such that 鈩�* = n鈧� + 鈩�* - 1: Since 鈩�* is the least element of T*, we know that for any element x in T*, 鈩�* <= x. So, 鈩�* <= (鈩�* - 1 + n鈧�), which means that 鈩�* <= n鈧� + 鈩�* - 1. Since 鈩�* is an element of T*, there exists an element in T (let it be n鈥�) such that n鈥� - n鈧� + 1 = 鈩�*. Solving this equation for n鈥� we find that n鈥� = n鈧� + 鈩�* - 1. Since 鈩�* <= (鈩� - n鈧� + 1), we have n鈥� >= 鈩� for all 鈩� in S. Hence, n鈧� + 鈩�* - 1 is an element of S.

Now, we want to show that for any integer s in S, s is greater or equal to n鈧� + 鈩�* - 1. Let s be any element in S. Since S is a subset of T, s 鈮� n鈧�. We also know that s - n鈧� + 1 is an element of T* because T* = {n - n鈧� + 1 | n 鈭� T}. So, (s - n鈧� + 1) 鈭� T*. Since 鈩�* is the least element of T*, we know that 鈩�* <= (s - n鈧� + 1). Adding n鈧� - 1 to both sides of the inequality, we obtain n鈧� + 鈩�* - 1 <= s. Hence, s is greater or equal to n鈧� + 鈩�* - 1 for all s in S. Thus, n鈧� + 鈩�* - 1 is the least element of S.