Prove that union of countable sets is countable
$begingroup$
Could anyone explain to me the part in red? I can't see how the existence of the set T is used in the proof, and how theorem 2.8 is applied.
Here are the relevant definitions and theorem.
real-analysis proof-explanation
asked Dec 12 '18 at 18:55
Josh NgJosh Ng
977
$endgroup$
add a comment |
$begingroup$
Could anyone explain to me the part in red? I can't see how the existence of the set T is used in the proof, and how theorem 2.8 is applied.
Here are the relevant definitions and theorem.
real-analysis proof-explanation
asked Dec 12 '18 at 18:55
Josh NgJosh Ng
977
$endgroup$
$begingroup$
Suppose the sequence of elements is $1,2,2,3,4,4,4,6,,,,,,,$, then to each element we assign a label, i.e. the first element is $1$, the second is $2$, the third is also $2$ and so on. But when you make up the set $S$, you won't be repeating elements. So $S$ will have number 1, number 2, number 4, number 5, number 8... and so on. So the set $T={1,2,4,5,8, ldots} subset Bbb{N}$ is in one-one correspondence with $S$.
$endgroup$
– Anurag A
Dec 12 '18 at 19:05
add a comment |
$begingroup$
Could anyone explain to me the part in red? I can't see how the existence of the set T is used in the proof, and how theorem 2.8 is applied.
Here are the relevant definitions and theorem.
real-analysis proof-explanation
asked Dec 12 '18 at 18:55
Josh NgJosh Ng
977
$endgroup$
Could anyone explain to me the part in red? I can't see how the existence of the set T is used in the proof, and how theorem 2.8 is applied.
Here are the relevant definitions and theorem.
real-analysis proof-explanation
real-analysis proof-explanation
asked Dec 12 '18 at 18:55
Josh NgJosh Ng
977
asked Dec 12 '18 at 18:55
Josh NgJosh Ng
977
asked Dec 12 '18 at 18:55
Josh NgJosh Ng
977
asked Dec 12 '18 at 18:55
Josh NgJosh Ng
977
asked Dec 12 '18 at 18:55
Josh NgJosh Ng
977
977
$begingroup$
Suppose the sequence of elements is $1,2,2,3,4,4,4,6,,,,,,,$, then to each element we assign a label, i.e. the first element is $1$, the second is $2$, the third is also $2$ and so on. But when you make up the set $S$, you won't be repeating elements. So $S$ will have number 1, number 2, number 4, number 5, number 8... and so on. So the set $T={1,2,4,5,8, ldots} subset Bbb{N}$ is in one-one correspondence with $S$.
$endgroup$
– Anurag A
Dec 12 '18 at 19:05
add a comment |
$begingroup$
Suppose the sequence of elements is $1,2,2,3,4,4,4,6,,,,,,,$, then to each element we assign a label, i.e. the first element is $1$, the second is $2$, the third is also $2$ and so on. But when you make up the set $S$, you won't be repeating elements. So $S$ will have number 1, number 2, number 4, number 5, number 8... and so on. So the set $T={1,2,4,5,8, ldots} subset Bbb{N}$ is in one-one correspondence with $S$.
$endgroup$
– Anurag A
Dec 12 '18 at 19:05
$begingroup$
Suppose the sequence of elements is $1,2,2,3,4,4,4,6,,,,,,,$, then to each element we assign a label, i.e. the first element is $1$, the second is $2$, the third is also $2$ and so on. But when you make up the set $S$, you won't be repeating elements. So $S$ will have number 1, number 2, number 4, number 5, number 8... and so on. So the set $T={1,2,4,5,8, ldots} subset Bbb{N}$ is in one-one correspondence with $S$.
$endgroup$
– Anurag A
Dec 12 '18 at 19:05
$begingroup$
Suppose the sequence of elements is $1,2,2,3,4,4,4,6,,,,,,,$, then to each element we assign a label, i.e. the first element is $1$, the second is $2$, the third is also $2$ and so on. But when you make up the set $S$, you won't be repeating elements. So $S$ will have number 1, number 2, number 4, number 5, number 8... and so on. So the set $T={1,2,4,5,8, ldots} subset Bbb{N}$ is in one-one correspondence with $S$.
$endgroup$
– Anurag A
Dec 12 '18 at 19:05
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Following the arrows you get a surjective function $f:mathbb Ntobigcup_{n=1}^infty E_n$, but there might be repetitions, so it is not injective. Well then you just discard the duplicates and get a bijective function
$f:Tsubseteqmathbb Ntobigcup_{n=1}^infty E_n$.
answered Dec 12 '18 at 19:04
FedericoFederico
5,064514
$endgroup$
add a comment |
$begingroup$
The sequence is a function from $mathbb{N} rightarrow$ S, where $i mapsto x_i$.. This function is surjective by construction. Since there may be double counting, there is a subset T $subset mathbb{N}$ such that T $sim$ S.
answered Dec 12 '18 at 19:10
Joel PereiraJoel Pereira
75919
$endgroup$
add a comment |
$begingroup$
What you have in (17) is a function $g:mathbb Ntimesmathbb Nto S$. Namely, $g(k,j)=x_{k,j}$. You have that $g$ is surjective, but it may not be injective if there are elements repeated among the $E_n$.
Because $g$ is surjective, for each $sin S$ there exists $(x_1,y_1)inmathbb Ntimesmathbb N$ with $g(s_1,t_1)=s$. These pairs may not be unique if $g$ is not injective, but we may choose a single one for each $s$. Say $g(x_s,y_s)=s$. Now let
$$
T={(x_s,y_s): sin S}.
$$
Then $Ssim T$ and $Tsubset mathbb Ntimesmathbb N$. So $T$ is countable and $S$ is countable.
answered Dec 12 '18 at 19:06
Martin ArgeramiMartin Argerami
127k1182182
$endgroup$
$begingroup$
I was not talking about your answer, but rather the way it's phrased in the book, which might not be very clear for someone who is new to this.
$endgroup$
– Martin Argerami
Dec 12 '18 at 19:15
$begingroup$
Yes I agree that it is not sufficiently rigorous for something which should be quite pedagogical
$endgroup$
– Federico
Dec 12 '18 at 19:23
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Following the arrows you get a surjective function $f:mathbb Ntobigcup_{n=1}^infty E_n$, but there might be repetitions, so it is not injective. Well then you just discard the duplicates and get a bijective function
$f:Tsubseteqmathbb Ntobigcup_{n=1}^infty E_n$.
answered Dec 12 '18 at 19:04
FedericoFederico
5,064514
$endgroup$
add a comment |
$begingroup$
Following the arrows you get a surjective function $f:mathbb Ntobigcup_{n=1}^infty E_n$, but there might be repetitions, so it is not injective. Well then you just discard the duplicates and get a bijective function
$f:Tsubseteqmathbb Ntobigcup_{n=1}^infty E_n$.
answered Dec 12 '18 at 19:04
FedericoFederico
5,064514
$endgroup$
add a comment |
$begingroup$
Following the arrows you get a surjective function $f:mathbb Ntobigcup_{n=1}^infty E_n$, but there might be repetitions, so it is not injective. Well then you just discard the duplicates and get a bijective function
$f:Tsubseteqmathbb Ntobigcup_{n=1}^infty E_n$.
answered Dec 12 '18 at 19:04
FedericoFederico
5,064514
$endgroup$
Following the arrows you get a surjective function $f:mathbb Ntobigcup_{n=1}^infty E_n$, but there might be repetitions, so it is not injective. Well then you just discard the duplicates and get a bijective function
$f:Tsubseteqmathbb Ntobigcup_{n=1}^infty E_n$.
answered Dec 12 '18 at 19:04
FedericoFederico
5,064514
answered Dec 12 '18 at 19:04
FedericoFederico
5,064514
answered Dec 12 '18 at 19:04
FedericoFederico
5,064514
answered Dec 12 '18 at 19:04
FedericoFederico
5,064514
5,064514
add a comment |
add a comment |
$begingroup$
The sequence is a function from $mathbb{N} rightarrow$ S, where $i mapsto x_i$.. This function is surjective by construction. Since there may be double counting, there is a subset T $subset mathbb{N}$ such that T $sim$ S.
answered Dec 12 '18 at 19:10
Joel PereiraJoel Pereira
75919
$endgroup$
add a comment |
$begingroup$
The sequence is a function from $mathbb{N} rightarrow$ S, where $i mapsto x_i$.. This function is surjective by construction. Since there may be double counting, there is a subset T $subset mathbb{N}$ such that T $sim$ S.
answered Dec 12 '18 at 19:10
Joel PereiraJoel Pereira
75919
$endgroup$
add a comment |
$begingroup$
The sequence is a function from $mathbb{N} rightarrow$ S, where $i mapsto x_i$.. This function is surjective by construction. Since there may be double counting, there is a subset T $subset mathbb{N}$ such that T $sim$ S.
answered Dec 12 '18 at 19:10
Joel PereiraJoel Pereira
75919
$endgroup$
The sequence is a function from $mathbb{N} rightarrow$ S, where $i mapsto x_i$.. This function is surjective by construction. Since there may be double counting, there is a subset T $subset mathbb{N}$ such that T $sim$ S.
answered Dec 12 '18 at 19:10
Joel PereiraJoel Pereira
75919
answered Dec 12 '18 at 19:10
Joel PereiraJoel Pereira
75919
answered Dec 12 '18 at 19:10
Joel PereiraJoel Pereira
75919
answered Dec 12 '18 at 19:10
Joel PereiraJoel Pereira
75919
75919
add a comment |
add a comment |
$begingroup$
What you have in (17) is a function $g:mathbb Ntimesmathbb Nto S$. Namely, $g(k,j)=x_{k,j}$. You have that $g$ is surjective, but it may not be injective if there are elements repeated among the $E_n$.
Because $g$ is surjective, for each $sin S$ there exists $(x_1,y_1)inmathbb Ntimesmathbb N$ with $g(s_1,t_1)=s$. These pairs may not be unique if $g$ is not injective, but we may choose a single one for each $s$. Say $g(x_s,y_s)=s$. Now let
$$
T={(x_s,y_s): sin S}.
$$
Then $Ssim T$ and $Tsubset mathbb Ntimesmathbb N$. So $T$ is countable and $S$ is countable.
answered Dec 12 '18 at 19:06
Martin ArgeramiMartin Argerami
127k1182182
$endgroup$
$begingroup$
I was not talking about your answer, but rather the way it's phrased in the book, which might not be very clear for someone who is new to this.
$endgroup$
– Martin Argerami
Dec 12 '18 at 19:15
$begingroup$
Yes I agree that it is not sufficiently rigorous for something which should be quite pedagogical
$endgroup$
– Federico
Dec 12 '18 at 19:23
add a comment |
$begingroup$
What you have in (17) is a function $g:mathbb Ntimesmathbb Nto S$. Namely, $g(k,j)=x_{k,j}$. You have that $g$ is surjective, but it may not be injective if there are elements repeated among the $E_n$.
Because $g$ is surjective, for each $sin S$ there exists $(x_1,y_1)inmathbb Ntimesmathbb N$ with $g(s_1,t_1)=s$. These pairs may not be unique if $g$ is not injective, but we may choose a single one for each $s$. Say $g(x_s,y_s)=s$. Now let
$$
T={(x_s,y_s): sin S}.
$$
Then $Ssim T$ and $Tsubset mathbb Ntimesmathbb N$. So $T$ is countable and $S$ is countable.
answered Dec 12 '18 at 19:06
Martin ArgeramiMartin Argerami
127k1182182
$endgroup$
$begingroup$
I was not talking about your answer, but rather the way it's phrased in the book, which might not be very clear for someone who is new to this.
$endgroup$
– Martin Argerami
Dec 12 '18 at 19:15
$begingroup$
Yes I agree that it is not sufficiently rigorous for something which should be quite pedagogical
$endgroup$
– Federico
Dec 12 '18 at 19:23
add a comment |
$begingroup$
What you have in (17) is a function $g:mathbb Ntimesmathbb Nto S$. Namely, $g(k,j)=x_{k,j}$. You have that $g$ is surjective, but it may not be injective if there are elements repeated among the $E_n$.
Because $g$ is surjective, for each $sin S$ there exists $(x_1,y_1)inmathbb Ntimesmathbb N$ with $g(s_1,t_1)=s$. These pairs may not be unique if $g$ is not injective, but we may choose a single one for each $s$. Say $g(x_s,y_s)=s$. Now let
$$
T={(x_s,y_s): sin S}.
$$
Then $Ssim T$ and $Tsubset mathbb Ntimesmathbb N$. So $T$ is countable and $S$ is countable.
answered Dec 12 '18 at 19:06
Martin ArgeramiMartin Argerami
127k1182182
$endgroup$
What you have in (17) is a function $g:mathbb Ntimesmathbb Nto S$. Namely, $g(k,j)=x_{k,j}$. You have that $g$ is surjective, but it may not be injective if there are elements repeated among the $E_n$.
Because $g$ is surjective, for each $sin S$ there exists $(x_1,y_1)inmathbb Ntimesmathbb N$ with $g(s_1,t_1)=s$. These pairs may not be unique if $g$ is not injective, but we may choose a single one for each $s$. Say $g(x_s,y_s)=s$. Now let
$$
T={(x_s,y_s): sin S}.
$$
Then $Ssim T$ and $Tsubset mathbb Ntimesmathbb N$. So $T$ is countable and $S$ is countable.
answered Dec 12 '18 at 19:06
Martin ArgeramiMartin Argerami
127k1182182
edited Dec 12 '18 at 19:14
answered Dec 12 '18 at 19:06
Martin ArgeramiMartin Argerami
127k1182182
answered Dec 12 '18 at 19:06
Martin ArgeramiMartin Argerami
127k1182182
answered Dec 12 '18 at 19:06
Martin ArgeramiMartin Argerami
127k1182182
127k1182182
$begingroup$
I was not talking about your answer, but rather the way it's phrased in the book, which might not be very clear for someone who is new to this.
$endgroup$
– Martin Argerami
Dec 12 '18 at 19:15
$begingroup$
Yes I agree that it is not sufficiently rigorous for something which should be quite pedagogical
$endgroup$
– Federico
Dec 12 '18 at 19:23
add a comment |
$begingroup$
I was not talking about your answer, but rather the way it's phrased in the book, which might not be very clear for someone who is new to this.
$endgroup$
– Martin Argerami
Dec 12 '18 at 19:15
$begingroup$
Yes I agree that it is not sufficiently rigorous for something which should be quite pedagogical
$endgroup$
– Federico
Dec 12 '18 at 19:23
$begingroup$
I was not talking about your answer, but rather the way it's phrased in the book, which might not be very clear for someone who is new to this.
$endgroup$
– Martin Argerami
Dec 12 '18 at 19:15
$begingroup$
I was not talking about your answer, but rather the way it's phrased in the book, which might not be very clear for someone who is new to this.
$endgroup$
– Martin Argerami
Dec 12 '18 at 19:15
$begingroup$
Yes I agree that it is not sufficiently rigorous for something which should be quite pedagogical
$endgroup$
– Federico
Dec 12 '18 at 19:23
$begingroup$
Yes I agree that it is not sufficiently rigorous for something which should be quite pedagogical
$endgroup$
– Federico
Dec 12 '18 at 19:23
add a comment |
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$begingroup$
Suppose the sequence of elements is $1,2,2,3,4,4,4,6,,,,,,,$, then to each element we assign a label, i.e. the first element is $1$, the second is $2$, the third is also $2$ and so on. But when you make up the set $S$, you won't be repeating elements. So $S$ will have number 1, number 2, number 4, number 5, number 8... and so on. So the set $T={1,2,4,5,8, ldots} subset Bbb{N}$ is in one-one correspondence with $S$.
$endgroup$
– Anurag A
Dec 12 '18 at 19:05