Finite and Infinite Sets. Countability. Proof Techniques
... So D cannot be anywhere among the rows of the table. However we have assumed that the table contains all possible subsets of N. This is a contradiction, following from our assumption, that the elements of 2N can be ordered. Hence 2N is uncountable. D. Direct Proof and Proof by Contradiction Example ...
... So D cannot be anywhere among the rows of the table. However we have assumed that the table contains all possible subsets of N. This is a contradiction, following from our assumption, that the elements of 2N can be ordered. Hence 2N is uncountable. D. Direct Proof and Proof by Contradiction Example ...
The Axiom of Choice
... order on the integers Z is not a well-ordering: choose any subset which is not bounded below, e.g. {0, −1, −2, . . .} (or for that matter all of Z), and such a subset will not have a least element. However, you can come up with a different total ordering of the integers which is a well-ordering; fo ...
... order on the integers Z is not a well-ordering: choose any subset which is not bounded below, e.g. {0, −1, −2, . . .} (or for that matter all of Z), and such a subset will not have a least element. However, you can come up with a different total ordering of the integers which is a well-ordering; fo ...
arXiv:math/0511682v1 [math.NT] 28 Nov 2005
... this question was first considered by Khintchine in [21] (see also [6, 38, 40] for surveys including a discussion on this subject). A preliminary step towards its resolution consists in providing explicit examples of transcendental continued fractions. The first result of this type goes back to the ...
... this question was first considered by Khintchine in [21] (see also [6, 38, 40] for surveys including a discussion on this subject). A preliminary step towards its resolution consists in providing explicit examples of transcendental continued fractions. The first result of this type goes back to the ...