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C(reals) = C(P(naturals))?? |
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| Nov26-03, 06:10 AM | #1 |
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C(reals) = C(P(naturals))??
hello,
Could someone help me please. I am studying Cantor's set theory at present, but am a little confused as to why he concludes that the cardinality of the set of real numbers is equal to the cardinal number of the power set of naturals (2^aleph0). Thanks. |
| Nov26-03, 11:43 AM | #2 |
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Recognitions:
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The elements of 2N can be treated as sequences of 1's and 0's where a sequence corresponding to a subset A of N has a 1 in the n th position if n is in A.
You should have no trouble seeing that this mapping is bijective. Now, if you look at the real numbers on [0,1] base 2, you get numbers like: 0.10101011101000110... which are also sequences of ones and zeros. So there's a natural mapping. Unfortunately there is a problem because 0.011111111111111111111111....= 0.100000000000000000000000.... in the reals. But that only occurs a countable number of (N) times. So we can certainly construct a bijection to [0,1] + N. So we have |2N| = |[0,1] + N| but you should already know that |[0,1] + N|=|[0,1]|=|R| (Cantor certainly did) So by substitution we get: |2N |=|R| |
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