Set of subsets containing individual elements
Consider the set $S={a,b,c}$. I am trying to formalize a general approach to fragmenting the set $S$, s.t. a fragmentation function $F$ is a function that maps from $S$ to $2^S$.
The granularity, of course depends on the specific function, e.g. the function $F_C(S)={forall t|tin S}$ is very coarse-granular that returns the set {{a,b,c}}.
However, how do I formalize a fragmentation function that returns the set {{a},{b},{c}}? The only idea I have for it is $F_F={forall{t}|tin S}$, however it just seems wrong. Is the function $F_C$ even correctly formulated?
Edit: My question is generally this: How do I formalize a function $F_F(S)$ that is injective and maps $a$ to ${a}$, $b$ to ${b}$ etc?
Thanks! - Christian
elementary-set-theory
asked Nov 29 '18 at 12:47
ChraebeChraebe
1032
|
show 3 more comments
Consider the set $S={a,b,c}$. I am trying to formalize a general approach to fragmenting the set $S$, s.t. a fragmentation function $F$ is a function that maps from $S$ to $2^S$.
The granularity, of course depends on the specific function, e.g. the function $F_C(S)={forall t|tin S}$ is very coarse-granular that returns the set {{a,b,c}}.
However, how do I formalize a fragmentation function that returns the set {{a},{b},{c}}? The only idea I have for it is $F_F={forall{t}|tin S}$, however it just seems wrong. Is the function $F_C$ even correctly formulated?
Edit: My question is generally this: How do I formalize a function $F_F(S)$ that is injective and maps $a$ to ${a}$, $b$ to ${b}$ etc?
Thanks! - Christian
elementary-set-theory
asked Nov 29 '18 at 12:47
ChraebeChraebe
1032
What's a fragmentation function? Specifically, what properties do you want your function $F$ to have?
– lulu
Nov 29 '18 at 12:57
@lulu - Essentially a function that creates a set of fragments, where a fragment $f$ is a subset of $S$, $fsubseteq S$, that is $F: Srightarrow 2^S$.
– Chraebe
Nov 29 '18 at 12:59
Not sure that clarifies anything. A function from $S$ to $2^S$ can only have at most three values in its range. I guess you can pick whatever three subsets you like. Are you sure you don't want the domain of $F$ to be the power set of $S$?
– lulu
Nov 29 '18 at 13:00
@lulu I thought the power set $S$ was $2^S$? But anyways, yes that is what I want
– Chraebe
Nov 29 '18 at 13:02
So, you want a function from the power set of $S$ to $2^S$? What properties would you like that function to have?
– lulu
Nov 29 '18 at 13:03
|
show 3 more comments
Consider the set $S={a,b,c}$. I am trying to formalize a general approach to fragmenting the set $S$, s.t. a fragmentation function $F$ is a function that maps from $S$ to $2^S$.
The granularity, of course depends on the specific function, e.g. the function $F_C(S)={forall t|tin S}$ is very coarse-granular that returns the set {{a,b,c}}.
However, how do I formalize a fragmentation function that returns the set {{a},{b},{c}}? The only idea I have for it is $F_F={forall{t}|tin S}$, however it just seems wrong. Is the function $F_C$ even correctly formulated?
Edit: My question is generally this: How do I formalize a function $F_F(S)$ that is injective and maps $a$ to ${a}$, $b$ to ${b}$ etc?
Thanks! - Christian
elementary-set-theory
asked Nov 29 '18 at 12:47
ChraebeChraebe
1032
Consider the set $S={a,b,c}$. I am trying to formalize a general approach to fragmenting the set $S$, s.t. a fragmentation function $F$ is a function that maps from $S$ to $2^S$.
The granularity, of course depends on the specific function, e.g. the function $F_C(S)={forall t|tin S}$ is very coarse-granular that returns the set {{a,b,c}}.
However, how do I formalize a fragmentation function that returns the set {{a},{b},{c}}? The only idea I have for it is $F_F={forall{t}|tin S}$, however it just seems wrong. Is the function $F_C$ even correctly formulated?
Edit: My question is generally this: How do I formalize a function $F_F(S)$ that is injective and maps $a$ to ${a}$, $b$ to ${b}$ etc?
Thanks! - Christian
elementary-set-theory
elementary-set-theory
asked Nov 29 '18 at 12:47
ChraebeChraebe
1032
asked Nov 29 '18 at 12:47
ChraebeChraebe
1032
edited Nov 29 '18 at 13:39
Chraebe
asked Nov 29 '18 at 12:47
ChraebeChraebe
1032
asked Nov 29 '18 at 12:47
ChraebeChraebe
1032
asked Nov 29 '18 at 12:47
ChraebeChraebe
1032
1032
What's a fragmentation function? Specifically, what properties do you want your function $F$ to have?
– lulu
Nov 29 '18 at 12:57
@lulu - Essentially a function that creates a set of fragments, where a fragment $f$ is a subset of $S$, $fsubseteq S$, that is $F: Srightarrow 2^S$.
– Chraebe
Nov 29 '18 at 12:59
Not sure that clarifies anything. A function from $S$ to $2^S$ can only have at most three values in its range. I guess you can pick whatever three subsets you like. Are you sure you don't want the domain of $F$ to be the power set of $S$?
– lulu
Nov 29 '18 at 13:00
@lulu I thought the power set $S$ was $2^S$? But anyways, yes that is what I want
– Chraebe
Nov 29 '18 at 13:02
So, you want a function from the power set of $S$ to $2^S$? What properties would you like that function to have?
– lulu
Nov 29 '18 at 13:03
|
show 3 more comments
What's a fragmentation function? Specifically, what properties do you want your function $F$ to have?
– lulu
Nov 29 '18 at 12:57
@lulu - Essentially a function that creates a set of fragments, where a fragment $f$ is a subset of $S$, $fsubseteq S$, that is $F: Srightarrow 2^S$.
– Chraebe
Nov 29 '18 at 12:59
Not sure that clarifies anything. A function from $S$ to $2^S$ can only have at most three values in its range. I guess you can pick whatever three subsets you like. Are you sure you don't want the domain of $F$ to be the power set of $S$?
– lulu
Nov 29 '18 at 13:00
@lulu I thought the power set $S$ was $2^S$? But anyways, yes that is what I want
– Chraebe
Nov 29 '18 at 13:02
So, you want a function from the power set of $S$ to $2^S$? What properties would you like that function to have?
– lulu
Nov 29 '18 at 13:03
What's a fragmentation function? Specifically, what properties do you want your function $F$ to have?
– lulu
Nov 29 '18 at 12:57
What's a fragmentation function? Specifically, what properties do you want your function $F$ to have?
– lulu
Nov 29 '18 at 12:57
@lulu - Essentially a function that creates a set of fragments, where a fragment $f$ is a subset of $S$, $fsubseteq S$, that is $F: Srightarrow 2^S$.
– Chraebe
Nov 29 '18 at 12:59
@lulu - Essentially a function that creates a set of fragments, where a fragment $f$ is a subset of $S$, $fsubseteq S$, that is $F: Srightarrow 2^S$.
– Chraebe
Nov 29 '18 at 12:59
Not sure that clarifies anything. A function from $S$ to $2^S$ can only have at most three values in its range. I guess you can pick whatever three subsets you like. Are you sure you don't want the domain of $F$ to be the power set of $S$?
– lulu
Nov 29 '18 at 13:00
Not sure that clarifies anything. A function from $S$ to $2^S$ can only have at most three values in its range. I guess you can pick whatever three subsets you like. Are you sure you don't want the domain of $F$ to be the power set of $S$?
– lulu
Nov 29 '18 at 13:00
@lulu I thought the power set $S$ was $2^S$? But anyways, yes that is what I want
– Chraebe
Nov 29 '18 at 13:02
@lulu I thought the power set $S$ was $2^S$? But anyways, yes that is what I want
– Chraebe
Nov 29 '18 at 13:02
So, you want a function from the power set of $S$ to $2^S$? What properties would you like that function to have?
– lulu
Nov 29 '18 at 13:03
So, you want a function from the power set of $S$ to $2^S$? What properties would you like that function to have?
– lulu
Nov 29 '18 at 13:03
|
show 3 more comments
2 Answers
2
active
oldest
votes
If $F:Stowp(S)$ is prescribed by $xmapsto{x}$ then the graph of $F$ can be recognized as the set (of ordered pairs):$${langle x,{x}ranglemid xin S}$$
However this function is only bijective if $S$ is empty or is a singleton.
It is injective though, and to make it bijective its codomain must be restricted to its image which is the set ${{x}mid xin S}$.
Then we are dealing with $F:Sto{{x}mid xin S}$ prescribed by $xmapsto{x}$.
This function has the same graph as the original one, but if concepts like "bijective" are at stake then functions cannot be identified with their graphs only.
I am not sure about my own understanding of the question. Does this help?
answered Nov 29 '18 at 13:29
drhabdrhab
98.5k544129
Thank you. I meant that the function $F$ is injective, not bijective (sorry about that). Is ${langle x,{x}rangle | xin S}$ then the correct formalization of $F$ given that any $x$ is mapped to ${x}$?
– Chraebe
Nov 29 '18 at 13:45
In set theory a function $f$ is a set of ordered pairs that has a special property: if $langle x,yrangle,langle x,zranglein f$ then $y=z$. That is the case for $F={langle x,{x}ranglemid xin S}$. So it is a function and it is prescribed by $xmapsto{x}$. I don't know what you mean exactly with "formalization".
– drhab
Nov 29 '18 at 13:52
I see, I think I just misunderstood the notation of defining functions :). Another question - is the function $F_C={langle x,{S}rangle}$ then correct notation for a function where all elements in $S$ maps to ${S}$?
– Chraebe
Nov 29 '18 at 14:01
No, it should be ${langle x,{S}ranglemid xin S}$.
– drhab
Nov 29 '18 at 14:07
add a comment |
If this is what you ask: My question is generally this: How do I formalize a function $F_F(S)$ that is injective and maps $a$ to ${a}$, $b$ to ${b}$ etc?
Let: $$S={a,b,c}$$
(Aside: the above explication of $S$ is useless, since the following set and function will hold for any set.)
And, $$A={{x}| xin S}$$ then,
$$F_F:Arightarrow S$$
$$F_F={(x,{x})|xin S land {x}in A;x=x}$$
Or to avoid ambiguity,
$$A={{y}| yin S}$$
$$F_F={(x,{y})|xin S land {y}in A;x=y}$$
Notice:
$$x_1rightarrow y_1$$
$$x_2rightarrow y_2$$
$$x_nrightarrow y_n$$
Therefore,
$$(forall xin S (exists yin A) ni f(x)={y})$$
$$(f(x_1)={y_1} land f(x_2)={y_1})implies x_1=x_2$$
Meaning the function is bijective.
answered Nov 29 '18 at 13:51
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
354114
add a comment |
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2 Answers
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2 Answers
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active
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oldest
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If $F:Stowp(S)$ is prescribed by $xmapsto{x}$ then the graph of $F$ can be recognized as the set (of ordered pairs):$${langle x,{x}ranglemid xin S}$$
However this function is only bijective if $S$ is empty or is a singleton.
It is injective though, and to make it bijective its codomain must be restricted to its image which is the set ${{x}mid xin S}$.
Then we are dealing with $F:Sto{{x}mid xin S}$ prescribed by $xmapsto{x}$.
This function has the same graph as the original one, but if concepts like "bijective" are at stake then functions cannot be identified with their graphs only.
I am not sure about my own understanding of the question. Does this help?
answered Nov 29 '18 at 13:29
drhabdrhab
98.5k544129
Thank you. I meant that the function $F$ is injective, not bijective (sorry about that). Is ${langle x,{x}rangle | xin S}$ then the correct formalization of $F$ given that any $x$ is mapped to ${x}$?
– Chraebe
Nov 29 '18 at 13:45
In set theory a function $f$ is a set of ordered pairs that has a special property: if $langle x,yrangle,langle x,zranglein f$ then $y=z$. That is the case for $F={langle x,{x}ranglemid xin S}$. So it is a function and it is prescribed by $xmapsto{x}$. I don't know what you mean exactly with "formalization".
– drhab
Nov 29 '18 at 13:52
I see, I think I just misunderstood the notation of defining functions :). Another question - is the function $F_C={langle x,{S}rangle}$ then correct notation for a function where all elements in $S$ maps to ${S}$?
– Chraebe
Nov 29 '18 at 14:01
No, it should be ${langle x,{S}ranglemid xin S}$.
– drhab
Nov 29 '18 at 14:07
add a comment |
If $F:Stowp(S)$ is prescribed by $xmapsto{x}$ then the graph of $F$ can be recognized as the set (of ordered pairs):$${langle x,{x}ranglemid xin S}$$
However this function is only bijective if $S$ is empty or is a singleton.
It is injective though, and to make it bijective its codomain must be restricted to its image which is the set ${{x}mid xin S}$.
Then we are dealing with $F:Sto{{x}mid xin S}$ prescribed by $xmapsto{x}$.
This function has the same graph as the original one, but if concepts like "bijective" are at stake then functions cannot be identified with their graphs only.
I am not sure about my own understanding of the question. Does this help?
answered Nov 29 '18 at 13:29
drhabdrhab
98.5k544129
Thank you. I meant that the function $F$ is injective, not bijective (sorry about that). Is ${langle x,{x}rangle | xin S}$ then the correct formalization of $F$ given that any $x$ is mapped to ${x}$?
– Chraebe
Nov 29 '18 at 13:45
In set theory a function $f$ is a set of ordered pairs that has a special property: if $langle x,yrangle,langle x,zranglein f$ then $y=z$. That is the case for $F={langle x,{x}ranglemid xin S}$. So it is a function and it is prescribed by $xmapsto{x}$. I don't know what you mean exactly with "formalization".
– drhab
Nov 29 '18 at 13:52
I see, I think I just misunderstood the notation of defining functions :). Another question - is the function $F_C={langle x,{S}rangle}$ then correct notation for a function where all elements in $S$ maps to ${S}$?
– Chraebe
Nov 29 '18 at 14:01
No, it should be ${langle x,{S}ranglemid xin S}$.
– drhab
Nov 29 '18 at 14:07
add a comment |
If $F:Stowp(S)$ is prescribed by $xmapsto{x}$ then the graph of $F$ can be recognized as the set (of ordered pairs):$${langle x,{x}ranglemid xin S}$$
However this function is only bijective if $S$ is empty or is a singleton.
It is injective though, and to make it bijective its codomain must be restricted to its image which is the set ${{x}mid xin S}$.
Then we are dealing with $F:Sto{{x}mid xin S}$ prescribed by $xmapsto{x}$.
This function has the same graph as the original one, but if concepts like "bijective" are at stake then functions cannot be identified with their graphs only.
I am not sure about my own understanding of the question. Does this help?
answered Nov 29 '18 at 13:29
drhabdrhab
98.5k544129
If $F:Stowp(S)$ is prescribed by $xmapsto{x}$ then the graph of $F$ can be recognized as the set (of ordered pairs):$${langle x,{x}ranglemid xin S}$$
However this function is only bijective if $S$ is empty or is a singleton.
It is injective though, and to make it bijective its codomain must be restricted to its image which is the set ${{x}mid xin S}$.
Then we are dealing with $F:Sto{{x}mid xin S}$ prescribed by $xmapsto{x}$.
This function has the same graph as the original one, but if concepts like "bijective" are at stake then functions cannot be identified with their graphs only.
I am not sure about my own understanding of the question. Does this help?
answered Nov 29 '18 at 13:29
drhabdrhab
98.5k544129
answered Nov 29 '18 at 13:29
drhabdrhab
98.5k544129
answered Nov 29 '18 at 13:29
drhabdrhab
98.5k544129
answered Nov 29 '18 at 13:29
drhabdrhab
98.5k544129
98.5k544129
Thank you. I meant that the function $F$ is injective, not bijective (sorry about that). Is ${langle x,{x}rangle | xin S}$ then the correct formalization of $F$ given that any $x$ is mapped to ${x}$?
– Chraebe
Nov 29 '18 at 13:45
In set theory a function $f$ is a set of ordered pairs that has a special property: if $langle x,yrangle,langle x,zranglein f$ then $y=z$. That is the case for $F={langle x,{x}ranglemid xin S}$. So it is a function and it is prescribed by $xmapsto{x}$. I don't know what you mean exactly with "formalization".
– drhab
Nov 29 '18 at 13:52
I see, I think I just misunderstood the notation of defining functions :). Another question - is the function $F_C={langle x,{S}rangle}$ then correct notation for a function where all elements in $S$ maps to ${S}$?
– Chraebe
Nov 29 '18 at 14:01
No, it should be ${langle x,{S}ranglemid xin S}$.
– drhab
Nov 29 '18 at 14:07
add a comment |
Thank you. I meant that the function $F$ is injective, not bijective (sorry about that). Is ${langle x,{x}rangle | xin S}$ then the correct formalization of $F$ given that any $x$ is mapped to ${x}$?
– Chraebe
Nov 29 '18 at 13:45
In set theory a function $f$ is a set of ordered pairs that has a special property: if $langle x,yrangle,langle x,zranglein f$ then $y=z$. That is the case for $F={langle x,{x}ranglemid xin S}$. So it is a function and it is prescribed by $xmapsto{x}$. I don't know what you mean exactly with "formalization".
– drhab
Nov 29 '18 at 13:52
I see, I think I just misunderstood the notation of defining functions :). Another question - is the function $F_C={langle x,{S}rangle}$ then correct notation for a function where all elements in $S$ maps to ${S}$?
– Chraebe
Nov 29 '18 at 14:01
No, it should be ${langle x,{S}ranglemid xin S}$.
– drhab
Nov 29 '18 at 14:07
Thank you. I meant that the function $F$ is injective, not bijective (sorry about that). Is ${langle x,{x}rangle | xin S}$ then the correct formalization of $F$ given that any $x$ is mapped to ${x}$?
– Chraebe
Nov 29 '18 at 13:45
Thank you. I meant that the function $F$ is injective, not bijective (sorry about that). Is ${langle x,{x}rangle | xin S}$ then the correct formalization of $F$ given that any $x$ is mapped to ${x}$?
– Chraebe
Nov 29 '18 at 13:45
In set theory a function $f$ is a set of ordered pairs that has a special property: if $langle x,yrangle,langle x,zranglein f$ then $y=z$. That is the case for $F={langle x,{x}ranglemid xin S}$. So it is a function and it is prescribed by $xmapsto{x}$. I don't know what you mean exactly with "formalization".
– drhab
Nov 29 '18 at 13:52
In set theory a function $f$ is a set of ordered pairs that has a special property: if $langle x,yrangle,langle x,zranglein f$ then $y=z$. That is the case for $F={langle x,{x}ranglemid xin S}$. So it is a function and it is prescribed by $xmapsto{x}$. I don't know what you mean exactly with "formalization".
– drhab
Nov 29 '18 at 13:52
I see, I think I just misunderstood the notation of defining functions :). Another question - is the function $F_C={langle x,{S}rangle}$ then correct notation for a function where all elements in $S$ maps to ${S}$?
– Chraebe
Nov 29 '18 at 14:01
I see, I think I just misunderstood the notation of defining functions :). Another question - is the function $F_C={langle x,{S}rangle}$ then correct notation for a function where all elements in $S$ maps to ${S}$?
– Chraebe
Nov 29 '18 at 14:01
No, it should be ${langle x,{S}ranglemid xin S}$.
– drhab
Nov 29 '18 at 14:07
No, it should be ${langle x,{S}ranglemid xin S}$.
– drhab
Nov 29 '18 at 14:07
add a comment |
If this is what you ask: My question is generally this: How do I formalize a function $F_F(S)$ that is injective and maps $a$ to ${a}$, $b$ to ${b}$ etc?
Let: $$S={a,b,c}$$
(Aside: the above explication of $S$ is useless, since the following set and function will hold for any set.)
And, $$A={{x}| xin S}$$ then,
$$F_F:Arightarrow S$$
$$F_F={(x,{x})|xin S land {x}in A;x=x}$$
Or to avoid ambiguity,
$$A={{y}| yin S}$$
$$F_F={(x,{y})|xin S land {y}in A;x=y}$$
Notice:
$$x_1rightarrow y_1$$
$$x_2rightarrow y_2$$
$$x_nrightarrow y_n$$
Therefore,
$$(forall xin S (exists yin A) ni f(x)={y})$$
$$(f(x_1)={y_1} land f(x_2)={y_1})implies x_1=x_2$$
Meaning the function is bijective.
answered Nov 29 '18 at 13:51
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
354114
add a comment |
If this is what you ask: My question is generally this: How do I formalize a function $F_F(S)$ that is injective and maps $a$ to ${a}$, $b$ to ${b}$ etc?
Let: $$S={a,b,c}$$
(Aside: the above explication of $S$ is useless, since the following set and function will hold for any set.)
And, $$A={{x}| xin S}$$ then,
$$F_F:Arightarrow S$$
$$F_F={(x,{x})|xin S land {x}in A;x=x}$$
Or to avoid ambiguity,
$$A={{y}| yin S}$$
$$F_F={(x,{y})|xin S land {y}in A;x=y}$$
Notice:
$$x_1rightarrow y_1$$
$$x_2rightarrow y_2$$
$$x_nrightarrow y_n$$
Therefore,
$$(forall xin S (exists yin A) ni f(x)={y})$$
$$(f(x_1)={y_1} land f(x_2)={y_1})implies x_1=x_2$$
Meaning the function is bijective.
answered Nov 29 '18 at 13:51
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
354114
add a comment |
If this is what you ask: My question is generally this: How do I formalize a function $F_F(S)$ that is injective and maps $a$ to ${a}$, $b$ to ${b}$ etc?
Let: $$S={a,b,c}$$
(Aside: the above explication of $S$ is useless, since the following set and function will hold for any set.)
And, $$A={{x}| xin S}$$ then,
$$F_F:Arightarrow S$$
$$F_F={(x,{x})|xin S land {x}in A;x=x}$$
Or to avoid ambiguity,
$$A={{y}| yin S}$$
$$F_F={(x,{y})|xin S land {y}in A;x=y}$$
Notice:
$$x_1rightarrow y_1$$
$$x_2rightarrow y_2$$
$$x_nrightarrow y_n$$
Therefore,
$$(forall xin S (exists yin A) ni f(x)={y})$$
$$(f(x_1)={y_1} land f(x_2)={y_1})implies x_1=x_2$$
Meaning the function is bijective.
answered Nov 29 '18 at 13:51
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
354114
If this is what you ask: My question is generally this: How do I formalize a function $F_F(S)$ that is injective and maps $a$ to ${a}$, $b$ to ${b}$ etc?
Let: $$S={a,b,c}$$
(Aside: the above explication of $S$ is useless, since the following set and function will hold for any set.)
And, $$A={{x}| xin S}$$ then,
$$F_F:Arightarrow S$$
$$F_F={(x,{x})|xin S land {x}in A;x=x}$$
Or to avoid ambiguity,
$$A={{y}| yin S}$$
$$F_F={(x,{y})|xin S land {y}in A;x=y}$$
Notice:
$$x_1rightarrow y_1$$
$$x_2rightarrow y_2$$
$$x_nrightarrow y_n$$
Therefore,
$$(forall xin S (exists yin A) ni f(x)={y})$$
$$(f(x_1)={y_1} land f(x_2)={y_1})implies x_1=x_2$$
Meaning the function is bijective.
answered Nov 29 '18 at 13:51
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
354114
edited Nov 29 '18 at 14:07
answered Nov 29 '18 at 13:51
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
354114
answered Nov 29 '18 at 13:51
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
354114
answered Nov 29 '18 at 13:51
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
354114
354114
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What's a fragmentation function? Specifically, what properties do you want your function $F$ to have?
– lulu
Nov 29 '18 at 12:57
@lulu - Essentially a function that creates a set of fragments, where a fragment $f$ is a subset of $S$, $fsubseteq S$, that is $F: Srightarrow 2^S$.
– Chraebe
Nov 29 '18 at 12:59
Not sure that clarifies anything. A function from $S$ to $2^S$ can only have at most three values in its range. I guess you can pick whatever three subsets you like. Are you sure you don't want the domain of $F$ to be the power set of $S$?
– lulu
Nov 29 '18 at 13:00
@lulu I thought the power set $S$ was $2^S$? But anyways, yes that is what I want
– Chraebe
Nov 29 '18 at 13:02
So, you want a function from the power set of $S$ to $2^S$? What properties would you like that function to have?
– lulu
Nov 29 '18 at 13:03