Are there countable infinte decision problems or uncountable infinite?
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4A decision problem on a alphabet (i.e. non-empty finite set) $\Sigma$ is some subset $P\subseteq \Sigma^$ (i.e. $P\in\mathcal P(\Sigma^)$). So it's uncountable (because $\mathcal P(\Sigma^*)$ is). – xavierm02 Jan 20 '17 at 14:23
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@xavierm02 Please post answers as answers, not as comments. – David Richerby Jan 20 '17 at 15:05
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Uncountable. In fact you can see a decision problem P as a language L (the words corresponds with the inputs that are accepted by P) on alphabet Σ. A language can be encoded by an infinite binary string generating all the strings in Σ* in lexical order and associating 1 if the current string belongs to L, 0 otherwise.
The set of infinite binary strings S is uncountable infinite. Let's assume it is countable infinite. Then we can find an association between natural numbers and the strings in S, and list them. But if we take the string composed of the complement of the bits on the diagonal, then we obtain a new infinite binary string which belongs to S but is not in the list. Contraddiction.
So the set of languages (and the set of decision problems) is uncountable infinite.
Ack.
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