Use the Schroder-Bernstein theorem to show that (0, 1) and [0, 1] have the same cardinality.
Solution:Step_1: In this problem we need to show that (0,1) and [0,1]have the same cardinality by using the schroder-Bernstein theorem.Schroder-Bernstein theorem :If A and B are sets with and then A = B.The term cardinality refers to the number of members in a set.Cardinality can be finite or infinite. One -to-one function: A function for which every element of the range of the function corresponds to exactly one element of the domain. Test for one -to-one functions : If f(a) = f(b) implies that a = b , then f is one-to-one.Step-2:Given sets are : (0, 1) and [0,1].Now we have to show that is one-to-one function.(0 , 1) means 0 < x< 1 , and [0 , 1] means .So, clearly (0,1) is a subset of [0,1] then there exist an one-to-one function such that f(x) = x , for all .If a = b , then f(a) = f(b) is true. Therefore , is one-to-one function. Step-3:Now we have to show that is one-to-one function.Consider , any interval between (0,1).For our convenience let us take [0.3, 0.7], here the interval length is 0.4.Clearly, .So, first interval is [0,0.4] , add 0.3 then the second interval is [0.3,0.7]So, g(x) is a function from [0 ,1] to [0.3,0.7] .Now, consider the function g(x) = (0.4)x+ 0.3 .Clearly, [0.3,0.7] is a subset of (0,1),then there exist an one-to-one function such that g(x) = (0.4)x+ 0.3 , for all .If a = b , then g(a) = g(b) is true.Consider,a = b Therefore , is one-to-one function.Step-4:Therefore , is one-to-one function and is one-to-one function.By Schroder-Bernstein theorem we can say that |(0,1) |=| [0,1]| Therefore , (0,1) and [0,1]have the same cardinality by using the schroder-Bernstein theorem.