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Definition 12.1.5. If two sets can be put into one-to-one correspondence then they are said to have the same cardinality. With these two definitions in place we can see that Theorem 12.1.3 is nothing less than the statement that the real numbers are not countably infinite. Since it is certainly not finite, then we say that the set of real numbers is uncountable and therefore "bigger" than the natural numbers! To see this let us suppose first that each real number appears in the sequence (sn)1 exactly once. In that case the indexing of our sequence is really just a one-to-one correspondence between the elements of the sequence and N : 1 → $1 2 + S2 3 + S3 4 → $4 If some real numbers are repeated in our sequence then all of the real numbers are a subset of our sequence and will therefore also be countable (see Problem 12.1.7, part a). Problem 12.1.7. Let {A;} be a collection of countable sets. Show that each of the following sets is also countable: (a) Any subset of A1. (b) A1 U A2 (c) A1 U A2 U A3 (d) UA; i=1 (e) UA; i=1 in-context In either case, every sequence is countable. But our theorem says that no sequence in R includes all of R. Therefore R is uncountable. Most of the sets you have encountered so far in your life have been countable.

Definition 12.1.5. If two sets can be put into one-to-one correspondence then they are said to have the same cardinality. With these two definitions in place we can see that Theorem 12.1.3 is nothing less than the statement that the real numbers are not countably infinite. Since it is certainly not finite, then we say that the set of real numbers is uncountable and therefore "bigger" than the natural numbers! To see this let us suppose first that each real number appears in the sequence (sn)1 exactly once. In that case the indexing of our sequence is really just a one-to-one correspondence between the elements of the sequence and N : 1 → $1 2 + S2 3 + S3 4 → $4 If some real numbers are repeated in our sequence then all of the real numbers are a subset of our sequence and will therefore also be countable (see Problem 12.1.7, part a). Problem 12.1.7. Let {A;} be a collection of countable sets. Show that each of the following sets is also countable: (a) Any subset of A1. (b) A1 U A2 (c) A1 U A2 U A3 (d) UA; i=1 (e) UA; i=1 in-context In either case, every sequence is countable. But our theorem says that no sequence in R includes all of R. Therefore R is uncountable. Most of the sets you have encountered so far in your life have been countable.

Algebra: Structure And Method, Book 1
Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter2: Working With Real Numbers
Section2.1: Basic Assumptions
Problem 40WE
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Real math analysis, Please help me solve problem 12.1.7

Definition 12.1.5. If two sets can be put into one-to-one correspondence then they are said to have the same cardinality. With these two definitions in place we can see that Theorem 12.1.3 is nothing less than the statement that the real numbers are not countably infinite. Since it is certainly not finite, then we say that the set of real numbers is uncountable and therefore "bigger" than the natural numbers! To see this let us suppose first that each real number appears in the sequence (sn)1 exactly once. In that case the indexing of our sequence is really just a one-to-one correspondence between the elements of the sequence and N : 1 → $1 2 + S2 3 + S3 4 → $4 If some real numbers are repeated in our sequence then all of the real numbers are a subset of our sequence and will therefore also be countable (see Problem 12.1.7, part a). Problem 12.1.7. Let {A;} be a collection of countable sets. Show that each of the following sets is also countable: (a) Any subset of A1. (b) A1 U A2 (c) A1 U A2 U A3 (d) UA; i=1 (e) UA; i=1 in-context In either case, every sequence is countable. But our theorem says that no sequence in R includes all of R. Therefore R is uncountable. Most of the sets you have encountered so far in your life have been countable.
Transcribed Image Text:Definition 12.1.5. If two sets can be put into one-to-one correspondence then they are said to have the same cardinality. With these two definitions in place we can see that Theorem 12.1.3 is nothing less than the statement that the real numbers are not countably infinite. Since it is certainly not finite, then we say that the set of real numbers is uncountable and therefore "bigger" than the natural numbers! To see this let us suppose first that each real number appears in the sequence (sn)1 exactly once. In that case the indexing of our sequence is really just a one-to-one correspondence between the elements of the sequence and N : 1 → $1 2 + S2 3 + S3 4 → $4 If some real numbers are repeated in our sequence then all of the real numbers are a subset of our sequence and will therefore also be countable (see Problem 12.1.7, part a). Problem 12.1.7. Let {A;} be a collection of countable sets. Show that each of the following sets is also countable: (a) Any subset of A1. (b) A1 U A2 (c) A1 U A2 U A3 (d) UA; i=1 (e) UA; i=1 in-context In either case, every sequence is countable. But our theorem says that no sequence in R includes all of R. Therefore R is uncountable. Most of the sets you have encountered so far in your life have been countable.
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