In elementary set theory, Cantor's theorem is a fundamental result which states that, for any set, the set of all subsets of (the power set of , denoted by ()) has a strictly greater cardinality than itself. set which is a contradiction. If $A$ is a finite set, then $|B|\leq |A| < \infty$, $$|W \cap B|=4$$ Furthermore, we designate the cardinality of countably infinite sets as ℵ0 ("aleph null"). Then U is an infinite subset of a countably infinite set A, so U is countably infinite by Theorem 14.8. << The second part of the theorem can be proved using the first part. This fact can be proved using a so-called diagonal argument, and we omit Thus by applying That is, there are 7 elements in the given set … If $A$ is countably infinite, then we can list the elements in $A$, To provide Without loss of generality, we may take $$A$$ = ℕ = {1, 2, 3,...}, the set of natural numbers. Before discussing }أ��-W�������"�C��� �00�!�U嚄l�'}�� -F�NQ gvPC���S�:|����ա���ʛ������v���|�Z���uo�2�aynް�K��gS��v�������*��P~Ē�����&63 that you can list the elements of a countable set $A$, i.e., you can write $A=\{a_1, a_2,\cdots\}$, $$|W \cup B \cup R|=21.$$ /Length 1933 more concrete, here we provide some useful results that help us prove if a set is countable or not. ?�iOQ���i�ʫ�.��_���׉!ә�. is concerned, this guideline should be sufficient for most cases. the proof here as it is not instrumental for the rest of the book. We have been able to create a list that contains all the elements in $\bigcup_{i} A_i$, so this ����{i�V�_�����A|%�v��{&F �B��oA�)QC|*i�P@c���$[B��X>�ʏ)+aK6���� -o��� �6� ;�I-#�a�F�*<9���*]����»n�s鿻摞���H���q��ѽ��n�WB_�S����c�ju�A:#�N���/u�,�0ki��2��:����!W�K/��H��'��Ym�R2n�)���2��;Û��&����:��'(��yt�Jzu�*Ĵ�1�&1}�yW7Q���m�M(���Q Ed ���ˀ������C�s� Ӌ��&�Qh��Ou���cJ����>���I6�'�/m��o��m�?R�"o�ͽP�����=�N��֩���&�5��y&���0 �$�YWs��M�ɵ{�ܘ.5Lθ�-� GL��sU 7����>��m�z������lW���)и�$0/�Z�P!�,r��VL�F��C�)�r�j�.F��|���Y_�p���P׍,�P��d�Oi��5'e��H���-cW_1TRg��LJ��q�(�GC�����7��Ps�b�\���U7��zM�d*1ɑ�]qV(�&3�&ޛtǸ"�^��6��Q|��|��_#�T� $$|A \cup B |=|A|+|B|-|A \cap B|.$$ refer to Figure 1.16 in Problem 2 to see this pictorially). According to the de nition, set has cardinality n when there is a sequence Thus, any set in this form is countable. Some sets that are not countable include ℝ, the set of real numbers between 0 and 1, and ℂ. Here is a simple guideline for deciding whether a set is countable or not.$\mathbb{N}, \mathbb{Z}, \mathbb{Q}$, and any of their subsets are countable. The Infinite Looper 48,403 views. but "bigger" sets such as$\mathbb{R}$are called uncountable. Thus U is both uncountable and countably infinite, a contradiction. Theorem13.1 Thereexistsabijection f :N!Z.Therefore jNj˘jZ. so it is an uncountable set. Becausethebijection f :N!Z matches up Nwith Z,itfollowsthat jj˘j.Wesummarizethiswithatheorem. In particular, one type is called countable, but you cannot list the elements in an uncountable set. /Length 1868 stream To see this, note that when we add$|A|$and$|B|$, we are counting the elements in$|A \cap B|$twice, thus$B$is countable. Maybe this is not so surprising, because N and Z have a strong geometric resemblance as sets of points on the number line. stream To be precise, here is the definition. If$A_1, A_2,\cdots$is a list of countable sets, then the set$\bigcup_{i} A_i=A_1 \cup A_2 \cup A_3\cdots$It is injective (“1 to 1”): f (x)=f (y) x=y. Thus according to Deﬁnition 2.3.1, the sets N and Z have the same cardinality. The second part of the theorem can be proved using the first part. S*~����7ׇ�E��bba&�Eo�oRB@3a͜9dQ�)ݶ�PSa�a�u��,�nP{|���Jq(jS�z1?m��h�^�aG?c��3>������1p+!��$�R��V�:�$��� �x�����2���/�d Figure 1.13 shows one possible ordering. Thus, �L�2��T�bg���H���g�-.n?�����|������xw���.�b6����,��,�fr��X��}ޖ�]uX��ՙ]�q�3����S���P7���W?s��c[u-���hEK�c��^�e�\�� If A is a finite set, then | B | ≤ | A | < ∞, thus B is countable. The proof of this theorem is very similar to the previous theorem. $$|W|=10$$ Cardinality (Screencast 5.1.4) ... Introduction to the Cardinality of Sets and a Countability Proof - Duration: 12:14. What if$A$is an infinite set? ����RJ�IR�� On the other hand, you cannot list the elements in$\mathbb{R}\$, If you are less interested in proofs, you may decide to skip them. If A is countably infinite, then we can list the elements in A, then by removing the elements in the list that are not in B, we can obtain a list for B, thus B is countable.

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