Mixing
![Counting number of words of length $n$ [closed]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
Baby Rudin's Proof in Theorem 2.13
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Theorem Let $A$ be a countable set, and let $B_n$ be the set of all n-tuples $(a_1, ..., a_n)$, where $a_k in A (k = 1, ..., n)$, and the elements $a_1, ..., a_n$ need not to be distinct. Then $B_n$ is countable.
BlockquoteProof That $B_1$ is countable is evident, since $B_1 = A$.(stop here)
Do I misunderstanding the definition of $B_n$?
In my opinion, set $B_1 =$ $(a_1, ..., a_n)$ , which means that there is merely one element in set $B_1$ which is an n-tuple $(a_1, ..., a_n)$. It is obvious that $B_1 neq A$. Would you please show some correct examples of $B_n$.
(continue) Suppose $B_n-1$ is countable $(n = 2, 3, 4, ...)$. The elements of $B_n$ are of the form
$$
(b,a) qquad (b in B_n-1, a in A).
$$
For every fixed $b$, the set of pairs $(b, a)$ is equivalent to $A$, and hence countable. Thus $B_n$ is the union of a countable set of countable sets. By Theorem 2.12, $B_n$ is countable.
And does $B_2$ = ({$a_1, ..., a_n), a_0)$ , $a_0 in A$ ?
Thanks in advance!
real-analysis
asked 2 days ago
lytoooo0
153
add a comment |Â
up vote
1
down vote
favorite
Theorem Let $A$ be a countable set, and let $B_n$ be the set of all n-tuples $(a_1, ..., a_n)$, where $a_k in A (k = 1, ..., n)$, and the elements $a_1, ..., a_n$ need not to be distinct. Then $B_n$ is countable.
BlockquoteProof That $B_1$ is countable is evident, since $B_1 = A$.(stop here)
Do I misunderstanding the definition of $B_n$?
In my opinion, set $B_1 =$ $(a_1, ..., a_n)$ , which means that there is merely one element in set $B_1$ which is an n-tuple $(a_1, ..., a_n)$. It is obvious that $B_1 neq A$. Would you please show some correct examples of $B_n$.
(continue) Suppose $B_n-1$ is countable $(n = 2, 3, 4, ...)$. The elements of $B_n$ are of the form
$$
(b,a) qquad (b in B_n-1, a in A).
$$
For every fixed $b$, the set of pairs $(b, a)$ is equivalent to $A$, and hence countable. Thus $B_n$ is the union of a countable set of countable sets. By Theorem 2.12, $B_n$ is countable.
And does $B_2$ = ({$a_1, ..., a_n), a_0)$ , $a_0 in A$ ?
Thanks in advance!
real-analysis
asked 2 days ago
lytoooo0
153
1
No, $B_1$ does not contain 1 $n$-tuple, contains all 1-tuples
– xarles
2 days ago
More simply, $B_n=A^n=Atimes Atimesdotstimes A$.
– Nicolas FRANCOIS
2 days ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Theorem Let $A$ be a countable set, and let $B_n$ be the set of all n-tuples $(a_1, ..., a_n)$, where $a_k in A (k = 1, ..., n)$, and the elements $a_1, ..., a_n$ need not to be distinct. Then $B_n$ is countable.
BlockquoteProof That $B_1$ is countable is evident, since $B_1 = A$.(stop here)
Do I misunderstanding the definition of $B_n$?
In my opinion, set $B_1 =$ $(a_1, ..., a_n)$ , which means that there is merely one element in set $B_1$ which is an n-tuple $(a_1, ..., a_n)$. It is obvious that $B_1 neq A$. Would you please show some correct examples of $B_n$.
(continue) Suppose $B_n-1$ is countable $(n = 2, 3, 4, ...)$. The elements of $B_n$ are of the form
$$
(b,a) qquad (b in B_n-1, a in A).
$$
For every fixed $b$, the set of pairs $(b, a)$ is equivalent to $A$, and hence countable. Thus $B_n$ is the union of a countable set of countable sets. By Theorem 2.12, $B_n$ is countable.
And does $B_2$ = ({$a_1, ..., a_n), a_0)$ , $a_0 in A$ ?
Thanks in advance!
real-analysis
asked 2 days ago
lytoooo0
153
Theorem Let $A$ be a countable set, and let $B_n$ be the set of all n-tuples $(a_1, ..., a_n)$, where $a_k in A (k = 1, ..., n)$, and the elements $a_1, ..., a_n$ need not to be distinct. Then $B_n$ is countable.
BlockquoteProof That $B_1$ is countable is evident, since $B_1 = A$.(stop here)
Do I misunderstanding the definition of $B_n$?
In my opinion, set $B_1 =$ $(a_1, ..., a_n)$ , which means that there is merely one element in set $B_1$ which is an n-tuple $(a_1, ..., a_n)$. It is obvious that $B_1 neq A$. Would you please show some correct examples of $B_n$.
(continue) Suppose $B_n-1$ is countable $(n = 2, 3, 4, ...)$. The elements of $B_n$ are of the form
$$
(b,a) qquad (b in B_n-1, a in A).
$$
For every fixed $b$, the set of pairs $(b, a)$ is equivalent to $A$, and hence countable. Thus $B_n$ is the union of a countable set of countable sets. By Theorem 2.12, $B_n$ is countable.
And does $B_2$ = ({$a_1, ..., a_n), a_0)$ , $a_0 in A$ ?
Thanks in advance!
real-analysis
asked 2 days ago
lytoooo0
153
asked 2 days ago
lytoooo0
153
asked 2 days ago
lytoooo0
153
asked 2 days ago
lytoooo0
153
153
1
No, $B_1$ does not contain 1 $n$-tuple, contains all 1-tuples
– xarles
2 days ago
More simply, $B_n=A^n=Atimes Atimesdotstimes A$.
– Nicolas FRANCOIS
2 days ago
add a comment |Â
1
No, $B_1$ does not contain 1 $n$-tuple, contains all 1-tuples
– xarles
2 days ago
More simply, $B_n=A^n=Atimes Atimesdotstimes A$.
– Nicolas FRANCOIS
2 days ago
1
1
No, $B_1$ does not contain 1 $n$-tuple, contains all 1-tuples
– xarles
2 days ago
No, $B_1$ does not contain 1 $n$-tuple, contains all 1-tuples
– xarles
2 days ago
More simply, $B_n=A^n=Atimes Atimesdotstimes A$.
– Nicolas FRANCOIS
2 days ago
More simply, $B_n=A^n=Atimes Atimesdotstimes A$.
– Nicolas FRANCOIS
2 days ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
$B_n$ is the set of all $n$-tuples of elements in $A$. Note that these are the same $n$.
So when $n=1$, $B_1$ is the set of all $1$-tuples of elements of $A$, ($B_1 = (a_1) : a_1 in A$) which we can identify with $A$.
When $n=2$ we have $B_2 = (a_1,a_2) : a_1, a_2 in A $, it's the set of all pairs of elements in $A$.
answered 2 days ago
Daniel Mroz
851314
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$B_n$ is the set of all $n$-tuples of elements in $A$. Note that these are the same $n$.
So when $n=1$, $B_1$ is the set of all $1$-tuples of elements of $A$, ($B_1 = (a_1) : a_1 in A$) which we can identify with $A$.
When $n=2$ we have $B_2 = (a_1,a_2) : a_1, a_2 in A $, it's the set of all pairs of elements in $A$.
answered 2 days ago
Daniel Mroz
851314
add a comment |Â
up vote
1
down vote
accepted
$B_n$ is the set of all $n$-tuples of elements in $A$. Note that these are the same $n$.
So when $n=1$, $B_1$ is the set of all $1$-tuples of elements of $A$, ($B_1 = (a_1) : a_1 in A$) which we can identify with $A$.
When $n=2$ we have $B_2 = (a_1,a_2) : a_1, a_2 in A $, it's the set of all pairs of elements in $A$.
answered 2 days ago
Daniel Mroz
851314
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$B_n$ is the set of all $n$-tuples of elements in $A$. Note that these are the same $n$.
So when $n=1$, $B_1$ is the set of all $1$-tuples of elements of $A$, ($B_1 = (a_1) : a_1 in A$) which we can identify with $A$.
When $n=2$ we have $B_2 = (a_1,a_2) : a_1, a_2 in A $, it's the set of all pairs of elements in $A$.
answered 2 days ago
Daniel Mroz
851314
$B_n$ is the set of all $n$-tuples of elements in $A$. Note that these are the same $n$.
So when $n=1$, $B_1$ is the set of all $1$-tuples of elements of $A$, ($B_1 = (a_1) : a_1 in A$) which we can identify with $A$.
When $n=2$ we have $B_2 = (a_1,a_2) : a_1, a_2 in A $, it's the set of all pairs of elements in $A$.
answered 2 days ago
Daniel Mroz
851314
answered 2 days ago
Daniel Mroz
851314
answered 2 days ago
Daniel Mroz
851314
answered 2 days ago
Daniel Mroz
851314
851314
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2871869%2fbaby-rudins-proof-in-theorem-2-13%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
No, $B_1$ does not contain 1 $n$-tuple, contains all 1-tuples
– xarles
2 days ago
More simply, $B_n=A^n=Atimes Atimesdotstimes A$.
– Nicolas FRANCOIS
2 days ago