Underlying set of the free monoid, does it contain the empty string?
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In the free monoid over a set the unique sequence of zero elements, often called the empty string is the identity element.
Is the empty string an element of the underlying set of the free monoid?
In the free monoid the elements of the underlying set are finite sequences of letters from an alphabet: $ {ab,aab,a,b...} $ I'm just wondering how exactly is the empty sequence represented in this underlying set?
Take a look at Wikipedia Kleene star:
If V is a set of symbols or characters, then $V*$ is the set of all strings over symbols in $V$, including the empty string $epsilon$.
category-theory monoid forgetful-functors
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In the free monoid over a set the unique sequence of zero elements, often called the empty string is the identity element.
Is the empty string an element of the underlying set of the free monoid?
In the free monoid the elements of the underlying set are finite sequences of letters from an alphabet: $ {ab,aab,a,b...} $ I'm just wondering how exactly is the empty sequence represented in this underlying set?
Take a look at Wikipedia Kleene star:
If V is a set of symbols or characters, then $V*$ is the set of all strings over symbols in $V$, including the empty string $epsilon$.
category-theory monoid forgetful-functors
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In the free monoid over a set the unique sequence of zero elements, often called the empty string is the identity element.
Is the empty string an element of the underlying set of the free monoid?
In the free monoid the elements of the underlying set are finite sequences of letters from an alphabet: $ {ab,aab,a,b...} $ I'm just wondering how exactly is the empty sequence represented in this underlying set?
Take a look at Wikipedia Kleene star:
If V is a set of symbols or characters, then $V*$ is the set of all strings over symbols in $V$, including the empty string $epsilon$.
category-theory monoid forgetful-functors
In the free monoid over a set the unique sequence of zero elements, often called the empty string is the identity element.
Is the empty string an element of the underlying set of the free monoid?
In the free monoid the elements of the underlying set are finite sequences of letters from an alphabet: $ {ab,aab,a,b...} $ I'm just wondering how exactly is the empty sequence represented in this underlying set?
Take a look at Wikipedia Kleene star:
If V is a set of symbols or characters, then $V*$ is the set of all strings over symbols in $V$, including the empty string $epsilon$.
category-theory monoid forgetful-functors
category-theory monoid forgetful-functors
edited Nov 20 at 14:53
asked Nov 20 at 10:54
Roland
19311
19311
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2 Answers
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A monoid is a tuple $(M,m,e)$ where $M$ is a set, $m:Mtimes M to M$ is an associative law and $ein M$ is a neutral element for $m$.
By definition, the underlying set of $(M,m,e)$ is $M$: therefore $ein M$ means that $e$ is an element of the underlying set.
In the free monoid generated by $A$, the neutral element is the empty string, so the empty string does belong to the underlying set of the free monoid, but there's nothing special about the free monoid here.
In the free monoid the elements of the underlying set are finite sequences of letters from an alphabet: ${ab, aab, a, b}$ I'm just wondering how exactly is the empty sequence represented in this underlying set.
– Roland
Nov 20 at 11:08
2
@Roland As the empty string. You can name it whatever you like.
– Tobias Kildetoft
Nov 20 at 11:17
3
@Roland Said another way, "finite" includes length 0!
– Kevin Carlson
Nov 20 at 17:21
add a comment |
up vote
3
down vote
Yes, it is.
If $M$ is the monoid free over set $S$ then you can identify the elements of $M$ with functions $nto S$ where $n$ is a nonnegative integer with $n:={0,dots,n-1}$.
Then $0$ is the empty set and the empty string is the empty function $0=varnothingto S$.
The empty function is also the empty set, so the empty string corresponds with $varnothing$.
If e.g. we are dealing with function $2to S$ of the form ${(0,a),(1,b)}$ then this function corresponds with string "ab".
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
A monoid is a tuple $(M,m,e)$ where $M$ is a set, $m:Mtimes M to M$ is an associative law and $ein M$ is a neutral element for $m$.
By definition, the underlying set of $(M,m,e)$ is $M$: therefore $ein M$ means that $e$ is an element of the underlying set.
In the free monoid generated by $A$, the neutral element is the empty string, so the empty string does belong to the underlying set of the free monoid, but there's nothing special about the free monoid here.
In the free monoid the elements of the underlying set are finite sequences of letters from an alphabet: ${ab, aab, a, b}$ I'm just wondering how exactly is the empty sequence represented in this underlying set.
– Roland
Nov 20 at 11:08
2
@Roland As the empty string. You can name it whatever you like.
– Tobias Kildetoft
Nov 20 at 11:17
3
@Roland Said another way, "finite" includes length 0!
– Kevin Carlson
Nov 20 at 17:21
add a comment |
up vote
2
down vote
accepted
A monoid is a tuple $(M,m,e)$ where $M$ is a set, $m:Mtimes M to M$ is an associative law and $ein M$ is a neutral element for $m$.
By definition, the underlying set of $(M,m,e)$ is $M$: therefore $ein M$ means that $e$ is an element of the underlying set.
In the free monoid generated by $A$, the neutral element is the empty string, so the empty string does belong to the underlying set of the free monoid, but there's nothing special about the free monoid here.
In the free monoid the elements of the underlying set are finite sequences of letters from an alphabet: ${ab, aab, a, b}$ I'm just wondering how exactly is the empty sequence represented in this underlying set.
– Roland
Nov 20 at 11:08
2
@Roland As the empty string. You can name it whatever you like.
– Tobias Kildetoft
Nov 20 at 11:17
3
@Roland Said another way, "finite" includes length 0!
– Kevin Carlson
Nov 20 at 17:21
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
A monoid is a tuple $(M,m,e)$ where $M$ is a set, $m:Mtimes M to M$ is an associative law and $ein M$ is a neutral element for $m$.
By definition, the underlying set of $(M,m,e)$ is $M$: therefore $ein M$ means that $e$ is an element of the underlying set.
In the free monoid generated by $A$, the neutral element is the empty string, so the empty string does belong to the underlying set of the free monoid, but there's nothing special about the free monoid here.
A monoid is a tuple $(M,m,e)$ where $M$ is a set, $m:Mtimes M to M$ is an associative law and $ein M$ is a neutral element for $m$.
By definition, the underlying set of $(M,m,e)$ is $M$: therefore $ein M$ means that $e$ is an element of the underlying set.
In the free monoid generated by $A$, the neutral element is the empty string, so the empty string does belong to the underlying set of the free monoid, but there's nothing special about the free monoid here.
answered Nov 20 at 11:03
Max
12.5k11040
12.5k11040
In the free monoid the elements of the underlying set are finite sequences of letters from an alphabet: ${ab, aab, a, b}$ I'm just wondering how exactly is the empty sequence represented in this underlying set.
– Roland
Nov 20 at 11:08
2
@Roland As the empty string. You can name it whatever you like.
– Tobias Kildetoft
Nov 20 at 11:17
3
@Roland Said another way, "finite" includes length 0!
– Kevin Carlson
Nov 20 at 17:21
add a comment |
In the free monoid the elements of the underlying set are finite sequences of letters from an alphabet: ${ab, aab, a, b}$ I'm just wondering how exactly is the empty sequence represented in this underlying set.
– Roland
Nov 20 at 11:08
2
@Roland As the empty string. You can name it whatever you like.
– Tobias Kildetoft
Nov 20 at 11:17
3
@Roland Said another way, "finite" includes length 0!
– Kevin Carlson
Nov 20 at 17:21
In the free monoid the elements of the underlying set are finite sequences of letters from an alphabet: ${ab, aab, a, b}$ I'm just wondering how exactly is the empty sequence represented in this underlying set.
– Roland
Nov 20 at 11:08
In the free monoid the elements of the underlying set are finite sequences of letters from an alphabet: ${ab, aab, a, b}$ I'm just wondering how exactly is the empty sequence represented in this underlying set.
– Roland
Nov 20 at 11:08
2
2
@Roland As the empty string. You can name it whatever you like.
– Tobias Kildetoft
Nov 20 at 11:17
@Roland As the empty string. You can name it whatever you like.
– Tobias Kildetoft
Nov 20 at 11:17
3
3
@Roland Said another way, "finite" includes length 0!
– Kevin Carlson
Nov 20 at 17:21
@Roland Said another way, "finite" includes length 0!
– Kevin Carlson
Nov 20 at 17:21
add a comment |
up vote
3
down vote
Yes, it is.
If $M$ is the monoid free over set $S$ then you can identify the elements of $M$ with functions $nto S$ where $n$ is a nonnegative integer with $n:={0,dots,n-1}$.
Then $0$ is the empty set and the empty string is the empty function $0=varnothingto S$.
The empty function is also the empty set, so the empty string corresponds with $varnothing$.
If e.g. we are dealing with function $2to S$ of the form ${(0,a),(1,b)}$ then this function corresponds with string "ab".
add a comment |
up vote
3
down vote
Yes, it is.
If $M$ is the monoid free over set $S$ then you can identify the elements of $M$ with functions $nto S$ where $n$ is a nonnegative integer with $n:={0,dots,n-1}$.
Then $0$ is the empty set and the empty string is the empty function $0=varnothingto S$.
The empty function is also the empty set, so the empty string corresponds with $varnothing$.
If e.g. we are dealing with function $2to S$ of the form ${(0,a),(1,b)}$ then this function corresponds with string "ab".
add a comment |
up vote
3
down vote
up vote
3
down vote
Yes, it is.
If $M$ is the monoid free over set $S$ then you can identify the elements of $M$ with functions $nto S$ where $n$ is a nonnegative integer with $n:={0,dots,n-1}$.
Then $0$ is the empty set and the empty string is the empty function $0=varnothingto S$.
The empty function is also the empty set, so the empty string corresponds with $varnothing$.
If e.g. we are dealing with function $2to S$ of the form ${(0,a),(1,b)}$ then this function corresponds with string "ab".
Yes, it is.
If $M$ is the monoid free over set $S$ then you can identify the elements of $M$ with functions $nto S$ where $n$ is a nonnegative integer with $n:={0,dots,n-1}$.
Then $0$ is the empty set and the empty string is the empty function $0=varnothingto S$.
The empty function is also the empty set, so the empty string corresponds with $varnothing$.
If e.g. we are dealing with function $2to S$ of the form ${(0,a),(1,b)}$ then this function corresponds with string "ab".
answered Nov 20 at 11:19
drhab
95.2k543126
95.2k543126
add a comment |
add a comment |
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