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.)

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
icon
Related questions
Question

Please help me with these questions. I am 

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.
Expert Solution
steps

Step by step

Solved in 3 steps

Blurred answer
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,