Answer:
Solution to determine whether each of these sets is countable or uncountable
Step-by-step explanation:
If A is countable then there exists an injective mapping f : A → Z+ which, for any S ⊆ A gives an injective mapping g : S → Z+ thereby establishing that S is countable. The contrapositive of this is: if a set is not countable then any superset is not countable. Â
(a) The rational numbers are countable (done in class) and this is a subset of the rational. Hence this set is also countable. Â
(b) this set is not countable. For contradiction suppose the elements of this set in (0,1) are enumerable. As in the diagonalization argument done in class we construct a number, r, in (0,1) whose decimal representation has as its i th digit (after the decimal) a digit different from the i th digit (after the decimal) of the i th number in the enumeration. Note that r can be constructed so that it does not have a 0 in its representation. Further, by construction r is different from all the other numbers in the enumeration thus yielding a contradiction