Cantor's diagonal argument: Difference between revisions
imported>Greg Woodhouse m (missed "to the") |
imported>Jitse Niesen (you need either S \subset N or S \in 2^N, but not S \subset 2^N; the domain of phi is N) |
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Line 3: | Line 3: | ||
==The Argument== | ==The Argument== | ||
To any set <math>S \subset | To any set <math>S \subset \mathbb{N}</math> we may associate a function <math>\phi : \mathbb{N} \rightarrow \{0, 1\}</math> by setting <math>\phi(n) = 1</math> if <math>n \in S</math> and <math>\phi(n) = 0</math>, otherwise. Conversely, every such function defines a subset. | ||
If power set is countable, there is a bijective map <math>\Psi : \mathbb{N} \rightarrow 2^{\mathbb{N}}</math>, that allows us to assign an index <math>i = \Psi^{-1} (S)</math> to every subset S. Assuming this has been done, we proceed to construct a function <math>\psi : \mathbb{N} \rightarrow \{ 0, 1\}</math> such that the corresponding set, <math>T</math> cannot be in the range of <math>\Psi</math>. | If power set is countable, there is a bijective map <math>\Psi : \mathbb{N} \rightarrow 2^{\mathbb{N}}</math>, that allows us to assign an index <math>i = \Psi^{-1} (S)</math> to every subset S. Assuming this has been done, we proceed to construct a function <math>\psi : \mathbb{N} \rightarrow \{ 0, 1\}</math> such that the corresponding set, <math>T</math> cannot be in the range of <math>\Psi</math>. |
Revision as of 07:42, 31 March 2007
Cantor's diagonal method provides a convenient proof that the set of subsets of the natural numbers (also known as its power set is not countable. More generally, it is a recurring theme in computability theory, where perhaps its most well known application is the negative solution to the halting problem.
The Argument
To any set we may associate a function by setting if and , otherwise. Conversely, every such function defines a subset.
If power set is countable, there is a bijective map , that allows us to assign an index to every subset S. Assuming this has been done, we proceed to construct a function such that the corresponding set, cannot be in the range of .
For each , either or , and so we may simply such that .
It follows that for any , and it must therefore correspond to a set not in the range of . This contradiction shows that cannot be countable.