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Identify the correct steps involved in proving that the union of a countable number of countable sets is countable. (Check all that apply.) Check All That Apply Since empty sets do not contribute any elements to unions, we can assume that none of the sets in our given countable collection of countable sets is an empty set. If there are no sets in the collection, then the union is empty and therefore countable. Otherwise let the countable sets be A₁, A2, .... Since each set A/is countable and nonempty, we can list its elements in a sequence as an, a2.... again, if the set is finite, we can list its elements and then list an repeatedly to assure an infinite sequence. Otherwise let the countable sets be A1, A2, .... Since each set A; is countable and nonempty, we cannot list its elements in a sequence as an, ap.... again, if the set is finite, we cannot list its elements and then list a repeatedly to assure an infinite sequence. We can put all the elements ajj into a sequence in a systematic way by listing all the elements aij in which i+j= 2 (there is only one such pair, (1, 1)), then all the elements in which i+j= 3 (there are only two such pairs, (1, 2) and (2, 1)), and so on; except that we do not list any element that we have already listed. We can put all the elements ajj into a sequence in a systematic way by listing all the elements aijin which i+j= 2 (there is only one such pair, (0, 0)), then all the elements in which i+j= 3 (there are only one such pair, (1, 2)), and so on; except that we do not list any element that we have already listed. So, assuming that these elements are distinct, our list starts a11, 12, a21. a13. a22. a31. a14..... (If any of these terms duplicates a previous term, then it is simply omitted.)

Identify the correct steps involved in proving that the union of a countable number of countable sets is countable. (Check all that apply.) Check All That Apply Since empty sets do not contribute any elements to unions, we can assume that none of the sets in our given countable collection of countable sets is an empty set. If there are no sets in the collection, then the union is empty and therefore countable. Otherwise let the countable sets be A₁, A2, .... Since each set A/is countable and nonempty, we can list its elements in a sequence as an, a2.... again, if the set is finite, we can list its elements and then list an repeatedly to assure an infinite sequence. Otherwise let the countable sets be A1, A2, .... Since each set A; is countable and nonempty, we cannot list its elements in a sequence as an, ap.... again, if the set is finite, we cannot list its elements and then list a repeatedly to assure an infinite sequence. We can put all the elements ajj into a sequence in a systematic way by listing all the elements aij in which i+j= 2 (there is only one such pair, (1, 1)), then all the elements in which i+j= 3 (there are only two such pairs, (1, 2) and (2, 1)), and so on; except that we do not list any element that we have already listed. We can put all the elements ajj into a sequence in a systematic way by listing all the elements aijin which i+j= 2 (there is only one such pair, (0, 0)), then all the elements in which i+j= 3 (there are only one such pair, (1, 2)), and so on; except that we do not list any element that we have already listed. So, assuming that these elements are distinct, our list starts a11, 12, a21. a13. a22. a31. a14..... (If any of these terms duplicates a previous term, then it is simply omitted.)

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter12: Probability
Section12.CR: Chapter 12 Review
Problem 4CR
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Identify the correct steps involved in proving that the union of a countable number of countable sets is countable. (Check all that apply.) Check All That Apply Since empty sets do not contribute any elements to unions, we can assume that none of the sets in our given countable collection of countable sets is an empty set. If there are no sets in the collection, then the union is empty and therefore countable. Otherwise let the countable sets be A1, A2, .... Since each set A; is countable and nonempty, we can list its elements in a sequence as an, a 12, .... again, if the set is finite, we can list its elements and then list an repeatedly to assure an infinite sequence. Otherwise let the countable sets be A₁, A2, .... Since each set Aj is countable and nonempty, we cannot list its elements in a sequence as an, a, ...; again, if the set is finite, we cannot list its elements and then list an repeatedly to assure an infinite sequence. We can put all the elements ajj into a sequence in a systematic way by listing all the elements aij in which i+j= 2 (there is only one such pair, (1, 1)), then all the elements in which i+j= 3 (there are only two such pairs, (1, 2) and (2, 1)), and so on; except that we do not list any element that we have already listed. We can put all the elements ajj into a sequence in a systematic way by listing all the elements ajj in which i+j= 2 (there is only one such pair, (0, 0)), then all the elements in which i+j= 3 (there are only one such pair, (1, 2)), and so on; except that we do not list any element that we have already listed. **** So, assuming that these elements are distinct, our list starts a11, 12, a21, a13, a22, 231, 14, .... (If any of these terms duplicates a previous term, then it is simply omitted.)
Transcribed Image Text:Identify the correct steps involved in proving that the union of a countable number of countable sets is countable. (Check all that apply.) Check All That Apply Since empty sets do not contribute any elements to unions, we can assume that none of the sets in our given countable collection of countable sets is an empty set. If there are no sets in the collection, then the union is empty and therefore countable. Otherwise let the countable sets be A1, A2, .... Since each set A; is countable and nonempty, we can list its elements in a sequence as an, a 12, .... again, if the set is finite, we can list its elements and then list an repeatedly to assure an infinite sequence. Otherwise let the countable sets be A₁, A2, .... Since each set Aj is countable and nonempty, we cannot list its elements in a sequence as an, a, ...; again, if the set is finite, we cannot list its elements and then list an repeatedly to assure an infinite sequence. We can put all the elements ajj into a sequence in a systematic way by listing all the elements aij in which i+j= 2 (there is only one such pair, (1, 1)), then all the elements in which i+j= 3 (there are only two such pairs, (1, 2) and (2, 1)), and so on; except that we do not list any element that we have already listed. We can put all the elements ajj into a sequence in a systematic way by listing all the elements ajj in which i+j= 2 (there is only one such pair, (0, 0)), then all the elements in which i+j= 3 (there are only one such pair, (1, 2)), and so on; except that we do not list any element that we have already listed. **** So, assuming that these elements are distinct, our list starts a11, 12, a21, a13, a22, 231, 14, .... (If any of these terms duplicates a previous term, then it is simply omitted.)
Arrange the steps in the correct order to prove the theorem "If A and B are sets, A is uncountable, and AC B, then B is uncountable." Rank the options below. The elements of B can be listed as b₁, b2, b3..... Assume that B is countable. Therefore A is countable, contradicting the hypothesis. Since A is a subset of B, taking the subsequence of [bn] that contains the terms that are in A gives a listing of the elements of A. Thus B is not countable.
Transcribed Image Text:Arrange the steps in the correct order to prove the theorem "If A and B are sets, A is uncountable, and AC B, then B is uncountable." Rank the options below. The elements of B can be listed as b₁, b2, b3..... Assume that B is countable. Therefore A is countable, contradicting the hypothesis. Since A is a subset of B, taking the subsequence of [bn] that contains the terms that are in A gives a listing of the elements of A. Thus B is not countable.
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