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It does this by showing that already the interval (0,1), i.e., the real numbers larger than 0 and smaller than 1, is not countably infinite. It proceeds as follows: |
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It does this by showing that already the interval (0,1), i.e., the real numbers larger than 0 and smaller than 1, is not countably infinite. It proceeds as follows: |
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: The digits we will consider are indicated in bold. From these digits we construct the digits of x as follows. |
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: The digits we will consider are indicated in bold. From these digits we define the digits of x as follows. |
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* (4) However, it differs in the nth decimal place from rn, so x is not in the set { r1, r2, r3, ... }. |
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* (4) However, it differs in the nth decimal place from rn, so x is not in the set { r1, r2, r3, ... }. |
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Note: Strictly speaking this argument only shows that the number of decimal expansions of real numbers between 0 and 1 is not countably infinite. But since there are expansions such as 0.01999... and 0.02000... that represent the same real number this does not trivially imply that the corresponding set of real numbers is also not countably infinite. This can be remedied by disallowing the decimal expansions that end with an infinite series of 9's. In that case it can be shown that for every real number there is a unique corresponding decimal expansion. It is easy to see that the proof then still proceeds because the number x contains only 1's and 0's in its decimal expansion. |
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Note: Strictly speaking, this argument only shows that the number of decimal expansions of real numbers between 0 and 1 is not countably infinite. But since there are expansions such as 0.01999... and 0.02000... that represent the same real number, this does not immediately imply that the corresponding set of real numbers is also not countably infinite. This can be remedied by disallowing the decimal expansions that end with an infinite series of 9's. In that case it can be shown that for every real number there is a unique corresponding decimal expansion. It is easy to see that the proof then still works because the number x contains only 1's and 0's in its decimal expansion. |
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* (2) Then there is a function f from S to P(S) that is surjective, i.e., for every V in P(S) there is an s in S such that f(s) = V. |
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* (2) Then there is a function f from S to P(S) that is surjective, i.e., for every V in P(S) there is an s in S such that f(s) = V. |
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