Concatenation operation on the set of finite sequences in ${0, 1}$
$begingroup$
Let $A^{ast} = bigcup_{I subset mathbb{N}} mathcal{F}(I, {0, 1}) = bigcup_{I subset mathbb{N}} (prod_{i in I} {0, 1})$ be the set of finite sequences in ${0, 1}$.
First, if $I = emptyset$, then we have $prod_{i in emptyset} {0, 1} = {u_{emptyset}} in A^{ast}$, where $u_{emptyset} : emptyset rightarrow {0, 1}$ is the empty mapping, right ?
Second, let $star$ be the concatenation operation on $A^{ast}$ defined in the following way :
$$forall (n, m) in mathbb{N}^{ast} times mathbb{N}^{ast}, forall u = (u_{1}, ..., u_{n}) in A^{ast}, forall v = (v_{1}, ..., v_{m}) in A^{ast}, u star v = (u_{1}, ..., u_{n}, v_{1}, ..., v_{m}) text{,}$$
I have seen somewhere that the neutral element for this operation is called the "empty sequence" (denoted here $()$) such that : $forall u in A^{ast}, u star () = () star u = u$. The problem is that I don't find a clear definition of what this empty sequence is.
Precisely, my questions are the following : Are the empty sequence and the empty mapping defined above in first point in fact the same element (i.e., $() = u_{emptyset}$) ? If it is not the case, what is exactly the empty sequence ?
Thank you for your help.
sequences-and-series binary-operations
asked Dec 8 '18 at 12:14
deeppinkwaterdeeppinkwater
658
$endgroup$
add a comment |
$begingroup$
Let $A^{ast} = bigcup_{I subset mathbb{N}} mathcal{F}(I, {0, 1}) = bigcup_{I subset mathbb{N}} (prod_{i in I} {0, 1})$ be the set of finite sequences in ${0, 1}$.
First, if $I = emptyset$, then we have $prod_{i in emptyset} {0, 1} = {u_{emptyset}} in A^{ast}$, where $u_{emptyset} : emptyset rightarrow {0, 1}$ is the empty mapping, right ?
Second, let $star$ be the concatenation operation on $A^{ast}$ defined in the following way :
$$forall (n, m) in mathbb{N}^{ast} times mathbb{N}^{ast}, forall u = (u_{1}, ..., u_{n}) in A^{ast}, forall v = (v_{1}, ..., v_{m}) in A^{ast}, u star v = (u_{1}, ..., u_{n}, v_{1}, ..., v_{m}) text{,}$$
I have seen somewhere that the neutral element for this operation is called the "empty sequence" (denoted here $()$) such that : $forall u in A^{ast}, u star () = () star u = u$. The problem is that I don't find a clear definition of what this empty sequence is.
Precisely, my questions are the following : Are the empty sequence and the empty mapping defined above in first point in fact the same element (i.e., $() = u_{emptyset}$) ? If it is not the case, what is exactly the empty sequence ?
Thank you for your help.
sequences-and-series binary-operations
asked Dec 8 '18 at 12:14
deeppinkwaterdeeppinkwater
658
$endgroup$
add a comment |
$begingroup$
Let $A^{ast} = bigcup_{I subset mathbb{N}} mathcal{F}(I, {0, 1}) = bigcup_{I subset mathbb{N}} (prod_{i in I} {0, 1})$ be the set of finite sequences in ${0, 1}$.
First, if $I = emptyset$, then we have $prod_{i in emptyset} {0, 1} = {u_{emptyset}} in A^{ast}$, where $u_{emptyset} : emptyset rightarrow {0, 1}$ is the empty mapping, right ?
Second, let $star$ be the concatenation operation on $A^{ast}$ defined in the following way :
$$forall (n, m) in mathbb{N}^{ast} times mathbb{N}^{ast}, forall u = (u_{1}, ..., u_{n}) in A^{ast}, forall v = (v_{1}, ..., v_{m}) in A^{ast}, u star v = (u_{1}, ..., u_{n}, v_{1}, ..., v_{m}) text{,}$$
I have seen somewhere that the neutral element for this operation is called the "empty sequence" (denoted here $()$) such that : $forall u in A^{ast}, u star () = () star u = u$. The problem is that I don't find a clear definition of what this empty sequence is.
Precisely, my questions are the following : Are the empty sequence and the empty mapping defined above in first point in fact the same element (i.e., $() = u_{emptyset}$) ? If it is not the case, what is exactly the empty sequence ?
Thank you for your help.
sequences-and-series binary-operations
asked Dec 8 '18 at 12:14
deeppinkwaterdeeppinkwater
658
$endgroup$
Let $A^{ast} = bigcup_{I subset mathbb{N}} mathcal{F}(I, {0, 1}) = bigcup_{I subset mathbb{N}} (prod_{i in I} {0, 1})$ be the set of finite sequences in ${0, 1}$.
First, if $I = emptyset$, then we have $prod_{i in emptyset} {0, 1} = {u_{emptyset}} in A^{ast}$, where $u_{emptyset} : emptyset rightarrow {0, 1}$ is the empty mapping, right ?
Second, let $star$ be the concatenation operation on $A^{ast}$ defined in the following way :
$$forall (n, m) in mathbb{N}^{ast} times mathbb{N}^{ast}, forall u = (u_{1}, ..., u_{n}) in A^{ast}, forall v = (v_{1}, ..., v_{m}) in A^{ast}, u star v = (u_{1}, ..., u_{n}, v_{1}, ..., v_{m}) text{,}$$
I have seen somewhere that the neutral element for this operation is called the "empty sequence" (denoted here $()$) such that : $forall u in A^{ast}, u star () = () star u = u$. The problem is that I don't find a clear definition of what this empty sequence is.
Precisely, my questions are the following : Are the empty sequence and the empty mapping defined above in first point in fact the same element (i.e., $() = u_{emptyset}$) ? If it is not the case, what is exactly the empty sequence ?
Thank you for your help.
sequences-and-series binary-operations
sequences-and-series binary-operations
asked Dec 8 '18 at 12:14
deeppinkwaterdeeppinkwater
658
asked Dec 8 '18 at 12:14
deeppinkwaterdeeppinkwater
658
asked Dec 8 '18 at 12:14
deeppinkwaterdeeppinkwater
658
asked Dec 8 '18 at 12:14
deeppinkwaterdeeppinkwater
658
asked Dec 8 '18 at 12:14
deeppinkwaterdeeppinkwater
658
658
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The OP's definition/setup is murky. Although there is no doubt about the intention, it is helpful to frame this with more precision. This is just one way to 'get formal'.
Definition: If $n in mathbb N$, then any function $u$ of the form
$tag 1 u: {k in mathbb N ; | ; 1 le k land k le n} to {0,1}$
is said to be a finite sequence in {0,1} of length $n$.
If $u$ is a finite sequence in {0,1} of length $n$ and $v$ is a finite sequence in {0,1} of length $m$ we define a finite sequence in {0,1} of length $n +m$, $u*v$, as follows:
$quadquadquadquad [u*v](k) = u(k) text{ for } k le n$
$quadquadquadquad [u*v](k) = v(k-n) text{ for } n + 1 lt k le n+m$
Informally, we write $u = (u_{1}, ..., u_{n})$.
The definition allows for a finite sequence of length $0$, but there can only be one form of such a sequence, the empty graph $emptyset$. In the same way that using logic shows that $0! = 1$, you can show that $emptyset$ serves as an identity.
Now if that sounds 'spooky', you can change the definition so that length $0$ is not allowed. Then, if you want, you can algebraically 'throw in' an identity with our associative binary operation of concatenation.
With our informal notation, using $()$ for $emptyset$ works great!
answered Dec 8 '18 at 14:38
CopyPasteItCopyPasteIt
4,1481628
$endgroup$
$begingroup$
In my case, the problem is that I need to allow sequences of length $0$ (so I'm agree, there's only one, the one I denoted "$u_{emptyset}$"), but I also need that when I do concatenation of any sequence $u ast u_{emptyset}$, I obtain $u$. So the algebraic identity for $ast$ can actually be $u_{emptyset}$ right ? Moreover (maybe I go to far sometimes, sorry), could we actually throw in another algebraic identity that is different from $u_{emptyset}$ (if it is one) or it is the only identity for $ast$ ?
$endgroup$
– deeppinkwater
Dec 8 '18 at 15:16
$begingroup$
@deeppinkwater You can throw in more than one algebraic identity and still have an associative operation - but why would you? When you put in just one identity, it make sense to say it has length $0$ - the length of a concatenation is the sum of the lengths.
$endgroup$
– CopyPasteIt
Dec 8 '18 at 15:32
$begingroup$
I don't want to throw another one. First, I need the existence of a sequence of length $0$ (without talking about concatenation), and then, I need an identity for concatenation and I want this identity to be the sequence of length $0$ I exhibited at first step (also, the problem in my case is that I don't really have an operation since I have a length $N$ that I cannot exceed, so concatenation is not really define for any sequence, but anyway...).
$endgroup$
– deeppinkwater
Dec 8 '18 at 15:41
$begingroup$
I'm asking you if I could put another identity because the one we talk about ($emptyset$) seems obvious to me and I don't realy see which one you could take instead of...
$endgroup$
– deeppinkwater
Dec 8 '18 at 15:41
$begingroup$
@deeppinkwater Any identity under multiplication (concatenation) does the same thing - nothing! So they all just look like the $emptyset$ Graph = Null Sequence. You can name it anything or use any notation you want. So use $text{ []} $ for the $emptyset$ Graph, and then $text{ []} * u = u * text{ []}$ for all $u$.
$endgroup$
– CopyPasteIt
Dec 9 '18 at 1:58
|
show 1 more comment
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1 Answer
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1 Answer
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$begingroup$
The OP's definition/setup is murky. Although there is no doubt about the intention, it is helpful to frame this with more precision. This is just one way to 'get formal'.
Definition: If $n in mathbb N$, then any function $u$ of the form
$tag 1 u: {k in mathbb N ; | ; 1 le k land k le n} to {0,1}$
is said to be a finite sequence in {0,1} of length $n$.
If $u$ is a finite sequence in {0,1} of length $n$ and $v$ is a finite sequence in {0,1} of length $m$ we define a finite sequence in {0,1} of length $n +m$, $u*v$, as follows:
$quadquadquadquad [u*v](k) = u(k) text{ for } k le n$
$quadquadquadquad [u*v](k) = v(k-n) text{ for } n + 1 lt k le n+m$
Informally, we write $u = (u_{1}, ..., u_{n})$.
The definition allows for a finite sequence of length $0$, but there can only be one form of such a sequence, the empty graph $emptyset$. In the same way that using logic shows that $0! = 1$, you can show that $emptyset$ serves as an identity.
Now if that sounds 'spooky', you can change the definition so that length $0$ is not allowed. Then, if you want, you can algebraically 'throw in' an identity with our associative binary operation of concatenation.
With our informal notation, using $()$ for $emptyset$ works great!
answered Dec 8 '18 at 14:38
CopyPasteItCopyPasteIt
4,1481628
$endgroup$
$begingroup$
In my case, the problem is that I need to allow sequences of length $0$ (so I'm agree, there's only one, the one I denoted "$u_{emptyset}$"), but I also need that when I do concatenation of any sequence $u ast u_{emptyset}$, I obtain $u$. So the algebraic identity for $ast$ can actually be $u_{emptyset}$ right ? Moreover (maybe I go to far sometimes, sorry), could we actually throw in another algebraic identity that is different from $u_{emptyset}$ (if it is one) or it is the only identity for $ast$ ?
$endgroup$
– deeppinkwater
Dec 8 '18 at 15:16
$begingroup$
@deeppinkwater You can throw in more than one algebraic identity and still have an associative operation - but why would you? When you put in just one identity, it make sense to say it has length $0$ - the length of a concatenation is the sum of the lengths.
$endgroup$
– CopyPasteIt
Dec 8 '18 at 15:32
$begingroup$
I don't want to throw another one. First, I need the existence of a sequence of length $0$ (without talking about concatenation), and then, I need an identity for concatenation and I want this identity to be the sequence of length $0$ I exhibited at first step (also, the problem in my case is that I don't really have an operation since I have a length $N$ that I cannot exceed, so concatenation is not really define for any sequence, but anyway...).
$endgroup$
– deeppinkwater
Dec 8 '18 at 15:41
$begingroup$
I'm asking you if I could put another identity because the one we talk about ($emptyset$) seems obvious to me and I don't realy see which one you could take instead of...
$endgroup$
– deeppinkwater
Dec 8 '18 at 15:41
$begingroup$
@deeppinkwater Any identity under multiplication (concatenation) does the same thing - nothing! So they all just look like the $emptyset$ Graph = Null Sequence. You can name it anything or use any notation you want. So use $text{ []} $ for the $emptyset$ Graph, and then $text{ []} * u = u * text{ []}$ for all $u$.
$endgroup$
– CopyPasteIt
Dec 9 '18 at 1:58
|
show 1 more comment
$begingroup$
The OP's definition/setup is murky. Although there is no doubt about the intention, it is helpful to frame this with more precision. This is just one way to 'get formal'.
Definition: If $n in mathbb N$, then any function $u$ of the form
$tag 1 u: {k in mathbb N ; | ; 1 le k land k le n} to {0,1}$
is said to be a finite sequence in {0,1} of length $n$.
If $u$ is a finite sequence in {0,1} of length $n$ and $v$ is a finite sequence in {0,1} of length $m$ we define a finite sequence in {0,1} of length $n +m$, $u*v$, as follows:
$quadquadquadquad [u*v](k) = u(k) text{ for } k le n$
$quadquadquadquad [u*v](k) = v(k-n) text{ for } n + 1 lt k le n+m$
Informally, we write $u = (u_{1}, ..., u_{n})$.
The definition allows for a finite sequence of length $0$, but there can only be one form of such a sequence, the empty graph $emptyset$. In the same way that using logic shows that $0! = 1$, you can show that $emptyset$ serves as an identity.
Now if that sounds 'spooky', you can change the definition so that length $0$ is not allowed. Then, if you want, you can algebraically 'throw in' an identity with our associative binary operation of concatenation.
With our informal notation, using $()$ for $emptyset$ works great!
answered Dec 8 '18 at 14:38
CopyPasteItCopyPasteIt
4,1481628
$endgroup$
$begingroup$
In my case, the problem is that I need to allow sequences of length $0$ (so I'm agree, there's only one, the one I denoted "$u_{emptyset}$"), but I also need that when I do concatenation of any sequence $u ast u_{emptyset}$, I obtain $u$. So the algebraic identity for $ast$ can actually be $u_{emptyset}$ right ? Moreover (maybe I go to far sometimes, sorry), could we actually throw in another algebraic identity that is different from $u_{emptyset}$ (if it is one) or it is the only identity for $ast$ ?
$endgroup$
– deeppinkwater
Dec 8 '18 at 15:16
$begingroup$
@deeppinkwater You can throw in more than one algebraic identity and still have an associative operation - but why would you? When you put in just one identity, it make sense to say it has length $0$ - the length of a concatenation is the sum of the lengths.
$endgroup$
– CopyPasteIt
Dec 8 '18 at 15:32
$begingroup$
I don't want to throw another one. First, I need the existence of a sequence of length $0$ (without talking about concatenation), and then, I need an identity for concatenation and I want this identity to be the sequence of length $0$ I exhibited at first step (also, the problem in my case is that I don't really have an operation since I have a length $N$ that I cannot exceed, so concatenation is not really define for any sequence, but anyway...).
$endgroup$
– deeppinkwater
Dec 8 '18 at 15:41
$begingroup$
I'm asking you if I could put another identity because the one we talk about ($emptyset$) seems obvious to me and I don't realy see which one you could take instead of...
$endgroup$
– deeppinkwater
Dec 8 '18 at 15:41
$begingroup$
@deeppinkwater Any identity under multiplication (concatenation) does the same thing - nothing! So they all just look like the $emptyset$ Graph = Null Sequence. You can name it anything or use any notation you want. So use $text{ []} $ for the $emptyset$ Graph, and then $text{ []} * u = u * text{ []}$ for all $u$.
$endgroup$
– CopyPasteIt
Dec 9 '18 at 1:58
|
show 1 more comment
$begingroup$
The OP's definition/setup is murky. Although there is no doubt about the intention, it is helpful to frame this with more precision. This is just one way to 'get formal'.
Definition: If $n in mathbb N$, then any function $u$ of the form
$tag 1 u: {k in mathbb N ; | ; 1 le k land k le n} to {0,1}$
is said to be a finite sequence in {0,1} of length $n$.
If $u$ is a finite sequence in {0,1} of length $n$ and $v$ is a finite sequence in {0,1} of length $m$ we define a finite sequence in {0,1} of length $n +m$, $u*v$, as follows:
$quadquadquadquad [u*v](k) = u(k) text{ for } k le n$
$quadquadquadquad [u*v](k) = v(k-n) text{ for } n + 1 lt k le n+m$
Informally, we write $u = (u_{1}, ..., u_{n})$.
The definition allows for a finite sequence of length $0$, but there can only be one form of such a sequence, the empty graph $emptyset$. In the same way that using logic shows that $0! = 1$, you can show that $emptyset$ serves as an identity.
Now if that sounds 'spooky', you can change the definition so that length $0$ is not allowed. Then, if you want, you can algebraically 'throw in' an identity with our associative binary operation of concatenation.
With our informal notation, using $()$ for $emptyset$ works great!
answered Dec 8 '18 at 14:38
CopyPasteItCopyPasteIt
4,1481628
$endgroup$
The OP's definition/setup is murky. Although there is no doubt about the intention, it is helpful to frame this with more precision. This is just one way to 'get formal'.
Definition: If $n in mathbb N$, then any function $u$ of the form
$tag 1 u: {k in mathbb N ; | ; 1 le k land k le n} to {0,1}$
is said to be a finite sequence in {0,1} of length $n$.
If $u$ is a finite sequence in {0,1} of length $n$ and $v$ is a finite sequence in {0,1} of length $m$ we define a finite sequence in {0,1} of length $n +m$, $u*v$, as follows:
$quadquadquadquad [u*v](k) = u(k) text{ for } k le n$
$quadquadquadquad [u*v](k) = v(k-n) text{ for } n + 1 lt k le n+m$
Informally, we write $u = (u_{1}, ..., u_{n})$.
The definition allows for a finite sequence of length $0$, but there can only be one form of such a sequence, the empty graph $emptyset$. In the same way that using logic shows that $0! = 1$, you can show that $emptyset$ serves as an identity.
Now if that sounds 'spooky', you can change the definition so that length $0$ is not allowed. Then, if you want, you can algebraically 'throw in' an identity with our associative binary operation of concatenation.
With our informal notation, using $()$ for $emptyset$ works great!
answered Dec 8 '18 at 14:38
CopyPasteItCopyPasteIt
4,1481628
edited Dec 8 '18 at 14:49
answered Dec 8 '18 at 14:38
CopyPasteItCopyPasteIt
4,1481628
answered Dec 8 '18 at 14:38
CopyPasteItCopyPasteIt
4,1481628
answered Dec 8 '18 at 14:38
CopyPasteItCopyPasteIt
4,1481628
4,1481628
$begingroup$
In my case, the problem is that I need to allow sequences of length $0$ (so I'm agree, there's only one, the one I denoted "$u_{emptyset}$"), but I also need that when I do concatenation of any sequence $u ast u_{emptyset}$, I obtain $u$. So the algebraic identity for $ast$ can actually be $u_{emptyset}$ right ? Moreover (maybe I go to far sometimes, sorry), could we actually throw in another algebraic identity that is different from $u_{emptyset}$ (if it is one) or it is the only identity for $ast$ ?
$endgroup$
– deeppinkwater
Dec 8 '18 at 15:16
$begingroup$
@deeppinkwater You can throw in more than one algebraic identity and still have an associative operation - but why would you? When you put in just one identity, it make sense to say it has length $0$ - the length of a concatenation is the sum of the lengths.
$endgroup$
– CopyPasteIt
Dec 8 '18 at 15:32
$begingroup$
I don't want to throw another one. First, I need the existence of a sequence of length $0$ (without talking about concatenation), and then, I need an identity for concatenation and I want this identity to be the sequence of length $0$ I exhibited at first step (also, the problem in my case is that I don't really have an operation since I have a length $N$ that I cannot exceed, so concatenation is not really define for any sequence, but anyway...).
$endgroup$
– deeppinkwater
Dec 8 '18 at 15:41
$begingroup$
I'm asking you if I could put another identity because the one we talk about ($emptyset$) seems obvious to me and I don't realy see which one you could take instead of...
$endgroup$
– deeppinkwater
Dec 8 '18 at 15:41
$begingroup$
@deeppinkwater Any identity under multiplication (concatenation) does the same thing - nothing! So they all just look like the $emptyset$ Graph = Null Sequence. You can name it anything or use any notation you want. So use $text{ []} $ for the $emptyset$ Graph, and then $text{ []} * u = u * text{ []}$ for all $u$.
$endgroup$
– CopyPasteIt
Dec 9 '18 at 1:58
|
show 1 more comment
$begingroup$
In my case, the problem is that I need to allow sequences of length $0$ (so I'm agree, there's only one, the one I denoted "$u_{emptyset}$"), but I also need that when I do concatenation of any sequence $u ast u_{emptyset}$, I obtain $u$. So the algebraic identity for $ast$ can actually be $u_{emptyset}$ right ? Moreover (maybe I go to far sometimes, sorry), could we actually throw in another algebraic identity that is different from $u_{emptyset}$ (if it is one) or it is the only identity for $ast$ ?
$endgroup$
– deeppinkwater
Dec 8 '18 at 15:16
$begingroup$
@deeppinkwater You can throw in more than one algebraic identity and still have an associative operation - but why would you? When you put in just one identity, it make sense to say it has length $0$ - the length of a concatenation is the sum of the lengths.
$endgroup$
– CopyPasteIt
Dec 8 '18 at 15:32
$begingroup$
I don't want to throw another one. First, I need the existence of a sequence of length $0$ (without talking about concatenation), and then, I need an identity for concatenation and I want this identity to be the sequence of length $0$ I exhibited at first step (also, the problem in my case is that I don't really have an operation since I have a length $N$ that I cannot exceed, so concatenation is not really define for any sequence, but anyway...).
$endgroup$
– deeppinkwater
Dec 8 '18 at 15:41
$begingroup$
I'm asking you if I could put another identity because the one we talk about ($emptyset$) seems obvious to me and I don't realy see which one you could take instead of...
$endgroup$
– deeppinkwater
Dec 8 '18 at 15:41
$begingroup$
@deeppinkwater Any identity under multiplication (concatenation) does the same thing - nothing! So they all just look like the $emptyset$ Graph = Null Sequence. You can name it anything or use any notation you want. So use $text{ []} $ for the $emptyset$ Graph, and then $text{ []} * u = u * text{ []}$ for all $u$.
$endgroup$
– CopyPasteIt
Dec 9 '18 at 1:58
$begingroup$
In my case, the problem is that I need to allow sequences of length $0$ (so I'm agree, there's only one, the one I denoted "$u_{emptyset}$"), but I also need that when I do concatenation of any sequence $u ast u_{emptyset}$, I obtain $u$. So the algebraic identity for $ast$ can actually be $u_{emptyset}$ right ? Moreover (maybe I go to far sometimes, sorry), could we actually throw in another algebraic identity that is different from $u_{emptyset}$ (if it is one) or it is the only identity for $ast$ ?
$endgroup$
– deeppinkwater
Dec 8 '18 at 15:16
$begingroup$
In my case, the problem is that I need to allow sequences of length $0$ (so I'm agree, there's only one, the one I denoted "$u_{emptyset}$"), but I also need that when I do concatenation of any sequence $u ast u_{emptyset}$, I obtain $u$. So the algebraic identity for $ast$ can actually be $u_{emptyset}$ right ? Moreover (maybe I go to far sometimes, sorry), could we actually throw in another algebraic identity that is different from $u_{emptyset}$ (if it is one) or it is the only identity for $ast$ ?
$endgroup$
– deeppinkwater
Dec 8 '18 at 15:16
$begingroup$
@deeppinkwater You can throw in more than one algebraic identity and still have an associative operation - but why would you? When you put in just one identity, it make sense to say it has length $0$ - the length of a concatenation is the sum of the lengths.
$endgroup$
– CopyPasteIt
Dec 8 '18 at 15:32
$begingroup$
@deeppinkwater You can throw in more than one algebraic identity and still have an associative operation - but why would you? When you put in just one identity, it make sense to say it has length $0$ - the length of a concatenation is the sum of the lengths.
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– CopyPasteIt
Dec 8 '18 at 15:32
$begingroup$
I don't want to throw another one. First, I need the existence of a sequence of length $0$ (without talking about concatenation), and then, I need an identity for concatenation and I want this identity to be the sequence of length $0$ I exhibited at first step (also, the problem in my case is that I don't really have an operation since I have a length $N$ that I cannot exceed, so concatenation is not really define for any sequence, but anyway...).
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– deeppinkwater
Dec 8 '18 at 15:41
$begingroup$
I don't want to throw another one. First, I need the existence of a sequence of length $0$ (without talking about concatenation), and then, I need an identity for concatenation and I want this identity to be the sequence of length $0$ I exhibited at first step (also, the problem in my case is that I don't really have an operation since I have a length $N$ that I cannot exceed, so concatenation is not really define for any sequence, but anyway...).
$endgroup$
– deeppinkwater
Dec 8 '18 at 15:41
$begingroup$
I'm asking you if I could put another identity because the one we talk about ($emptyset$) seems obvious to me and I don't realy see which one you could take instead of...
$endgroup$
– deeppinkwater
Dec 8 '18 at 15:41
$begingroup$
I'm asking you if I could put another identity because the one we talk about ($emptyset$) seems obvious to me and I don't realy see which one you could take instead of...
$endgroup$
– deeppinkwater
Dec 8 '18 at 15:41
$begingroup$
@deeppinkwater Any identity under multiplication (concatenation) does the same thing - nothing! So they all just look like the $emptyset$ Graph = Null Sequence. You can name it anything or use any notation you want. So use $text{ []} $ for the $emptyset$ Graph, and then $text{ []} * u = u * text{ []}$ for all $u$.
$endgroup$
– CopyPasteIt
Dec 9 '18 at 1:58
$begingroup$
@deeppinkwater Any identity under multiplication (concatenation) does the same thing - nothing! So they all just look like the $emptyset$ Graph = Null Sequence. You can name it anything or use any notation you want. So use $text{ []} $ for the $emptyset$ Graph, and then $text{ []} * u = u * text{ []}$ for all $u$.
$endgroup$
– CopyPasteIt
Dec 9 '18 at 1:58
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