- #1
linuxux
- 133
- 0
Hello, I'm working from a book called 'Understanding Analysis' by Stephen Abbott and there is a question I'm not sure about:
Explain why induction cannot be used to prove part (ii) of Theorem 1.4.13 from part (i).
Theorem 1.4.13 says:
(i) If [tex]A_1 , A_2 , \ldots A_m[/tex] are each countable sets, then the union [tex] A_1 \cup A_2 \cup \ldots \cup A_m [/tex] is countable.
(ii) If [tex]A_n[/tex] is a countable set for each [tex]n \in \mathbf{N}[/tex], then [tex]\cup^{\infty}_{n=1} A_n[/tex] is countable.
I thought it would be helpful to note that in the previous problem, I had two sets [tex]A_1[/tex] and [tex]A_2[/tex], both of which are countable, and had to prove that their union was also countable. in that instance, a set [tex]B_1[/tex] was made by [tex]A_1 / A_2 [/tex] which made [tex]A_1 \cup B_1[/tex] equal to [tex]A_1 \cup A_2[/tex] but [tex]A_1[/tex] and [tex]B_1[/tex] were disjoint, which was required in order to have a 1-1 function.
-im not sure if i could prove it by contradiction like this: i can assume there is a single list of all the combined elements in the sets. in this list there are nested intervals, and because the list will be mapped to a function, we assume there are no repeated elements in the list. so i define the list so element [tex]x_k[/tex] does not belong to interval [tex]I_{k+1}[/tex] (i'm not sure if this qualifies as an induction). thus the intersection of intervals is an empty set, as it should be. but the nested interval property says that the intersection is non-empty, so we know the list is missing at least on element, so we can't use the inductive method to prove part (ii) of Theorem 1.4.13 from part (i).
Explain why induction cannot be used to prove part (ii) of Theorem 1.4.13 from part (i).
Theorem 1.4.13 says:
(i) If [tex]A_1 , A_2 , \ldots A_m[/tex] are each countable sets, then the union [tex] A_1 \cup A_2 \cup \ldots \cup A_m [/tex] is countable.
(ii) If [tex]A_n[/tex] is a countable set for each [tex]n \in \mathbf{N}[/tex], then [tex]\cup^{\infty}_{n=1} A_n[/tex] is countable.
I thought it would be helpful to note that in the previous problem, I had two sets [tex]A_1[/tex] and [tex]A_2[/tex], both of which are countable, and had to prove that their union was also countable. in that instance, a set [tex]B_1[/tex] was made by [tex]A_1 / A_2 [/tex] which made [tex]A_1 \cup B_1[/tex] equal to [tex]A_1 \cup A_2[/tex] but [tex]A_1[/tex] and [tex]B_1[/tex] were disjoint, which was required in order to have a 1-1 function.
-im not sure if i could prove it by contradiction like this: i can assume there is a single list of all the combined elements in the sets. in this list there are nested intervals, and because the list will be mapped to a function, we assume there are no repeated elements in the list. so i define the list so element [tex]x_k[/tex] does not belong to interval [tex]I_{k+1}[/tex] (i'm not sure if this qualifies as an induction). thus the intersection of intervals is an empty set, as it should be. but the nested interval property says that the intersection is non-empty, so we know the list is missing at least on element, so we can't use the inductive method to prove part (ii) of Theorem 1.4.13 from part (i).
Last edited: