Mixing
Countable set of sequences - is there a sequence where every element is greater or equal?
$begingroup$
I'm looking at this question:
$mathrm { N } ^ { mathrm { N } } : = { f : f : mathrm { N } rightarrow mathrm { N } }$ is the set of all sequences of natural numbers. Let $A= left{ f _ { n } in mathbb { N } ^ { mathrm { N } } : n in mathrm { N } right}$ be any countable subset (finite or infinite) of $mathrm { N } ^ { mathrm { N } }$. Is there a sequence $f in mathrm { N } ^ { mathrm { N } }$ where $f _ { n } leq ^ { * } fquad forall n in mathrm { N } $?
My assumption would be that the answer is yes, however I have no idea how to even begin to prove it. I'm guessing the fact that $A$ is countable is important. My only idea would be that, because it is countable, you can possibly order the sequences for each n and then let $f(n)$ be the element of the sequence which is ranked highest, but I really don't know.
I would be thankful for any suggestions!
Edit:
Definition of $fleq ^ { * } g$ : $exists m in mathbb { N } : forall n in mathbb { N } text { where } n geq m Longrightarrow f ( n ) leq g ( n )$
sequences-and-series set-theory
edited Jan 20 at 22:45
Noah Schweber
126k10151290
asked Jan 20 at 21:13
LHeartHLHeartH
83
$endgroup$
|
show 1 more comment
$begingroup$
I'm looking at this question:
$mathrm { N } ^ { mathrm { N } } : = { f : f : mathrm { N } rightarrow mathrm { N } }$ is the set of all sequences of natural numbers. Let $A= left{ f _ { n } in mathbb { N } ^ { mathrm { N } } : n in mathrm { N } right}$ be any countable subset (finite or infinite) of $mathrm { N } ^ { mathrm { N } }$. Is there a sequence $f in mathrm { N } ^ { mathrm { N } }$ where $f _ { n } leq ^ { * } fquad forall n in mathrm { N } $?
My assumption would be that the answer is yes, however I have no idea how to even begin to prove it. I'm guessing the fact that $A$ is countable is important. My only idea would be that, because it is countable, you can possibly order the sequences for each n and then let $f(n)$ be the element of the sequence which is ranked highest, but I really don't know.
I would be thankful for any suggestions!
Edit:
Definition of $fleq ^ { * } g$ : $exists m in mathbb { N } : forall n in mathbb { N } text { where } n geq m Longrightarrow f ( n ) leq g ( n )$
sequences-and-series set-theory
edited Jan 20 at 22:45
Noah Schweber
126k10151290
asked Jan 20 at 21:13
LHeartHLHeartH
83
$endgroup$
$begingroup$
By $fleq^*g$ do you mean $f(n) leq g(n) forall nin mathbb{N}$?
$endgroup$
– bruderjakob17
Jan 20 at 21:20
$begingroup$
Yes, sorry, that was defined elsewhere.
$endgroup$
– LHeartH
Jan 20 at 21:21
1
$begingroup$
That is not the usual definition of $le^*$. Please check what definition you are actually using.
$endgroup$
– Andrés E. Caicedo
Jan 20 at 21:22
$begingroup$
Sorry again... It was only vaguely defined in two sentences ahead of the question. I did however find this definition in my lecturer's notes later on: $exists m in mathbb { N } text { : } forall n in mathbb { N } text { where } n geq m implies f ( n ) leq g ( n )$
$endgroup$
– LHeartH
Jan 20 at 21:35
$begingroup$
Please edit that definition into your question. It is the more common definition. It ruins the answer you have gotten.
$endgroup$
– Ross Millikan
Jan 20 at 22:14
|
show 1 more comment
$begingroup$
I'm looking at this question:
$mathrm { N } ^ { mathrm { N } } : = { f : f : mathrm { N } rightarrow mathrm { N } }$ is the set of all sequences of natural numbers. Let $A= left{ f _ { n } in mathbb { N } ^ { mathrm { N } } : n in mathrm { N } right}$ be any countable subset (finite or infinite) of $mathrm { N } ^ { mathrm { N } }$. Is there a sequence $f in mathrm { N } ^ { mathrm { N } }$ where $f _ { n } leq ^ { * } fquad forall n in mathrm { N } $?
My assumption would be that the answer is yes, however I have no idea how to even begin to prove it. I'm guessing the fact that $A$ is countable is important. My only idea would be that, because it is countable, you can possibly order the sequences for each n and then let $f(n)$ be the element of the sequence which is ranked highest, but I really don't know.
I would be thankful for any suggestions!
Edit:
Definition of $fleq ^ { * } g$ : $exists m in mathbb { N } : forall n in mathbb { N } text { where } n geq m Longrightarrow f ( n ) leq g ( n )$
sequences-and-series set-theory
edited Jan 20 at 22:45
Noah Schweber
126k10151290
asked Jan 20 at 21:13
LHeartHLHeartH
83
$endgroup$
I'm looking at this question:
$mathrm { N } ^ { mathrm { N } } : = { f : f : mathrm { N } rightarrow mathrm { N } }$ is the set of all sequences of natural numbers. Let $A= left{ f _ { n } in mathbb { N } ^ { mathrm { N } } : n in mathrm { N } right}$ be any countable subset (finite or infinite) of $mathrm { N } ^ { mathrm { N } }$. Is there a sequence $f in mathrm { N } ^ { mathrm { N } }$ where $f _ { n } leq ^ { * } fquad forall n in mathrm { N } $?
My assumption would be that the answer is yes, however I have no idea how to even begin to prove it. I'm guessing the fact that $A$ is countable is important. My only idea would be that, because it is countable, you can possibly order the sequences for each n and then let $f(n)$ be the element of the sequence which is ranked highest, but I really don't know.
I would be thankful for any suggestions!
Edit:
Definition of $fleq ^ { * } g$ : $exists m in mathbb { N } : forall n in mathbb { N } text { where } n geq m Longrightarrow f ( n ) leq g ( n )$
sequences-and-series set-theory
sequences-and-series set-theory
edited Jan 20 at 22:45
Noah Schweber
126k10151290
asked Jan 20 at 21:13
LHeartHLHeartH
83
edited Jan 20 at 22:45
Noah Schweber
126k10151290
asked Jan 20 at 21:13
LHeartHLHeartH
83
edited Jan 20 at 22:45
Noah Schweber
126k10151290
edited Jan 20 at 22:45
Noah Schweber
126k10151290
edited Jan 20 at 22:45
Noah Schweber
126k10151290
126k10151290
asked Jan 20 at 21:13
LHeartHLHeartH
83
asked Jan 20 at 21:13
LHeartHLHeartH
83
asked Jan 20 at 21:13
LHeartHLHeartH
83
83
$begingroup$
By $fleq^*g$ do you mean $f(n) leq g(n) forall nin mathbb{N}$?
$endgroup$
– bruderjakob17
Jan 20 at 21:20
$begingroup$
Yes, sorry, that was defined elsewhere.
$endgroup$
– LHeartH
Jan 20 at 21:21
1
$begingroup$
That is not the usual definition of $le^*$. Please check what definition you are actually using.
$endgroup$
– Andrés E. Caicedo
Jan 20 at 21:22
$begingroup$
Sorry again... It was only vaguely defined in two sentences ahead of the question. I did however find this definition in my lecturer's notes later on: $exists m in mathbb { N } text { : } forall n in mathbb { N } text { where } n geq m implies f ( n ) leq g ( n )$
$endgroup$
– LHeartH
Jan 20 at 21:35
$begingroup$
Please edit that definition into your question. It is the more common definition. It ruins the answer you have gotten.
$endgroup$
– Ross Millikan
Jan 20 at 22:14
|
show 1 more comment
$begingroup$
By $fleq^*g$ do you mean $f(n) leq g(n) forall nin mathbb{N}$?
$endgroup$
– bruderjakob17
Jan 20 at 21:20
$begingroup$
Yes, sorry, that was defined elsewhere.
$endgroup$
– LHeartH
Jan 20 at 21:21
1
$begingroup$
That is not the usual definition of $le^*$. Please check what definition you are actually using.
$endgroup$
– Andrés E. Caicedo
Jan 20 at 21:22
$begingroup$
Sorry again... It was only vaguely defined in two sentences ahead of the question. I did however find this definition in my lecturer's notes later on: $exists m in mathbb { N } text { : } forall n in mathbb { N } text { where } n geq m implies f ( n ) leq g ( n )$
$endgroup$
– LHeartH
Jan 20 at 21:35
$begingroup$
Please edit that definition into your question. It is the more common definition. It ruins the answer you have gotten.
$endgroup$
– Ross Millikan
Jan 20 at 22:14
$begingroup$
By $fleq^*g$ do you mean $f(n) leq g(n) forall nin mathbb{N}$?
$endgroup$
– bruderjakob17
Jan 20 at 21:20
$begingroup$
By $fleq^*g$ do you mean $f(n) leq g(n) forall nin mathbb{N}$?
$endgroup$
– bruderjakob17
Jan 20 at 21:20
$begingroup$
Yes, sorry, that was defined elsewhere.
$endgroup$
– LHeartH
Jan 20 at 21:21
$begingroup$
Yes, sorry, that was defined elsewhere.
$endgroup$
– LHeartH
Jan 20 at 21:21
1
1
$begingroup$
That is not the usual definition of $le^*$. Please check what definition you are actually using.
$endgroup$
– Andrés E. Caicedo
Jan 20 at 21:22
$begingroup$
That is not the usual definition of $le^*$. Please check what definition you are actually using.
$endgroup$
– Andrés E. Caicedo
Jan 20 at 21:22
$begingroup$
Sorry again... It was only vaguely defined in two sentences ahead of the question. I did however find this definition in my lecturer's notes later on: $exists m in mathbb { N } text { : } forall n in mathbb { N } text { where } n geq m implies f ( n ) leq g ( n )$
$endgroup$
– LHeartH
Jan 20 at 21:35
$begingroup$
Sorry again... It was only vaguely defined in two sentences ahead of the question. I did however find this definition in my lecturer's notes later on: $exists m in mathbb { N } text { : } forall n in mathbb { N } text { where } n geq m implies f ( n ) leq g ( n )$
$endgroup$
– LHeartH
Jan 20 at 21:35
$begingroup$
Please edit that definition into your question. It is the more common definition. It ruins the answer you have gotten.
$endgroup$
– Ross Millikan
Jan 20 at 22:14
$begingroup$
Please edit that definition into your question. It is the more common definition. It ruins the answer you have gotten.
$endgroup$
– Ross Millikan
Jan 20 at 22:14
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Yes, given a countable $A$ there is a sequence $f$ where all of them are $le^*$ than $f$. We set $$f(1)=f_1(1)+1\
f(2)=max(f_1(2),f_2(2))+1\
f(3)=max(f_1(3),f_2(3),f_3(3))+1\
f(n)=max_{i=1}^n(f_i(n))+1$$
This is greater than the first sequence starting at $1$, greater than the second starting by $2$, greater than the $n^{th}$ starting by $n$. Each of the $max$ functions has a finite list of arguments, so is well defined. Diagonalization wins again.
answered Jan 21 at 0:08
Ross MillikanRoss Millikan
298k24200373
$endgroup$
$begingroup$
Thank you! Got it now :)
$endgroup$
– LHeartH
Jan 21 at 7:01
$begingroup$
To the proposer: Alternatively let $f(n)=1+sum_{j=1}^n f_j(n)$.
$endgroup$
– DanielWainfleet
Jan 21 at 8:24
add a comment |
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1 Answer
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$begingroup$
Yes, given a countable $A$ there is a sequence $f$ where all of them are $le^*$ than $f$. We set $$f(1)=f_1(1)+1\
f(2)=max(f_1(2),f_2(2))+1\
f(3)=max(f_1(3),f_2(3),f_3(3))+1\
f(n)=max_{i=1}^n(f_i(n))+1$$
This is greater than the first sequence starting at $1$, greater than the second starting by $2$, greater than the $n^{th}$ starting by $n$. Each of the $max$ functions has a finite list of arguments, so is well defined. Diagonalization wins again.
answered Jan 21 at 0:08
Ross MillikanRoss Millikan
298k24200373
$endgroup$
$begingroup$
Thank you! Got it now :)
$endgroup$
– LHeartH
Jan 21 at 7:01
$begingroup$
To the proposer: Alternatively let $f(n)=1+sum_{j=1}^n f_j(n)$.
$endgroup$
– DanielWainfleet
Jan 21 at 8:24
add a comment |
$begingroup$
Yes, given a countable $A$ there is a sequence $f$ where all of them are $le^*$ than $f$. We set $$f(1)=f_1(1)+1\
f(2)=max(f_1(2),f_2(2))+1\
f(3)=max(f_1(3),f_2(3),f_3(3))+1\
f(n)=max_{i=1}^n(f_i(n))+1$$
This is greater than the first sequence starting at $1$, greater than the second starting by $2$, greater than the $n^{th}$ starting by $n$. Each of the $max$ functions has a finite list of arguments, so is well defined. Diagonalization wins again.
answered Jan 21 at 0:08
Ross MillikanRoss Millikan
298k24200373
$endgroup$
$begingroup$
Thank you! Got it now :)
$endgroup$
– LHeartH
Jan 21 at 7:01
$begingroup$
To the proposer: Alternatively let $f(n)=1+sum_{j=1}^n f_j(n)$.
$endgroup$
– DanielWainfleet
Jan 21 at 8:24
add a comment |
$begingroup$
Yes, given a countable $A$ there is a sequence $f$ where all of them are $le^*$ than $f$. We set $$f(1)=f_1(1)+1\
f(2)=max(f_1(2),f_2(2))+1\
f(3)=max(f_1(3),f_2(3),f_3(3))+1\
f(n)=max_{i=1}^n(f_i(n))+1$$
This is greater than the first sequence starting at $1$, greater than the second starting by $2$, greater than the $n^{th}$ starting by $n$. Each of the $max$ functions has a finite list of arguments, so is well defined. Diagonalization wins again.
answered Jan 21 at 0:08
Ross MillikanRoss Millikan
298k24200373
$endgroup$
Yes, given a countable $A$ there is a sequence $f$ where all of them are $le^*$ than $f$. We set $$f(1)=f_1(1)+1\
f(2)=max(f_1(2),f_2(2))+1\
f(3)=max(f_1(3),f_2(3),f_3(3))+1\
f(n)=max_{i=1}^n(f_i(n))+1$$
This is greater than the first sequence starting at $1$, greater than the second starting by $2$, greater than the $n^{th}$ starting by $n$. Each of the $max$ functions has a finite list of arguments, so is well defined. Diagonalization wins again.
answered Jan 21 at 0:08
Ross MillikanRoss Millikan
298k24200373
answered Jan 21 at 0:08
Ross MillikanRoss Millikan
298k24200373
answered Jan 21 at 0:08
Ross MillikanRoss Millikan
298k24200373
answered Jan 21 at 0:08
Ross MillikanRoss Millikan
298k24200373
298k24200373
$begingroup$
Thank you! Got it now :)
$endgroup$
– LHeartH
Jan 21 at 7:01
$begingroup$
To the proposer: Alternatively let $f(n)=1+sum_{j=1}^n f_j(n)$.
$endgroup$
– DanielWainfleet
Jan 21 at 8:24
add a comment |
$begingroup$
Thank you! Got it now :)
$endgroup$
– LHeartH
Jan 21 at 7:01
$begingroup$
To the proposer: Alternatively let $f(n)=1+sum_{j=1}^n f_j(n)$.
$endgroup$
– DanielWainfleet
Jan 21 at 8:24
$begingroup$
Thank you! Got it now :)
$endgroup$
– LHeartH
Jan 21 at 7:01
$begingroup$
Thank you! Got it now :)
$endgroup$
– LHeartH
Jan 21 at 7:01
$begingroup$
To the proposer: Alternatively let $f(n)=1+sum_{j=1}^n f_j(n)$.
$endgroup$
– DanielWainfleet
Jan 21 at 8:24
$begingroup$
To the proposer: Alternatively let $f(n)=1+sum_{j=1}^n f_j(n)$.
$endgroup$
– DanielWainfleet
Jan 21 at 8:24
add a comment |
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$begingroup$
By $fleq^*g$ do you mean $f(n) leq g(n) forall nin mathbb{N}$?
$endgroup$
– bruderjakob17
Jan 20 at 21:20
$begingroup$
Yes, sorry, that was defined elsewhere.
$endgroup$
– LHeartH
Jan 20 at 21:21
1
$begingroup$
That is not the usual definition of $le^*$. Please check what definition you are actually using.
$endgroup$
– Andrés E. Caicedo
Jan 20 at 21:22
$begingroup$
Sorry again... It was only vaguely defined in two sentences ahead of the question. I did however find this definition in my lecturer's notes later on: $exists m in mathbb { N } text { : } forall n in mathbb { N } text { where } n geq m implies f ( n ) leq g ( n )$
$endgroup$
– LHeartH
Jan 20 at 21:35
$begingroup$
Please edit that definition into your question. It is the more common definition. It ruins the answer you have gotten.
$endgroup$
– Ross Millikan
Jan 20 at 22:14