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Unformatted text preview: subsets of is not countable. answer: It is a proof by contradiction: i) there is a bijection between innite binary strings and subsets of , so we count the former instead; ii) assume, for a contradiction, that the set of innite binary strings is countable; iii) the strings can thus be arranged in a list s (1) , s (2) , s (3) , . . . . iv) Let b(i) be the complement of the i th bit of s ( i ) ; v) Then ~ b is a string which is not in the list, contradiction. 5. Show, using the pumping lemma for regular languages, that the set T consisting of binary string with more 1 s than 0 s is not regular. answer: By contradiction. Assume the set is regular and let be the pumping constant given by the pumping lemma. Let = 11 . Then T . The pumping lemma implies we can erase a (nonempty) substring from the prex 11 with the resulting string still being in T . But this clearly cannot be done, contradiction....
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This note was uploaded on 06/09/2011 for the course COT 4210 taught by Professor Staff during the Spring '08 term at University of Central Florida.
 Spring '08
 Staff
 Computer Science

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