Complete Lattice and Concept Lattice
$begingroup$
I am taking an online course of Introduction to Formal Concept Analysis and I'm trying to understand The Basic Theorem.
Well, if $mathscr{L}$ is a complete lattice, so $mathscr{L}cong underline{mathfrak{B}}(mathscr{L},mathscr{L},leq)$ (concept lattice of context $(mathscr{L},mathscr{L},leq)$).
And we know that:
(*) All concept lattice are complete lattice.
My question is: When we define a formal context $(G,M,I)$, the sets $G,M$ don't have 'conditions'. What happens when I take a non-complete lattice $mathscr{L}$ and define the formal context $(mathscr{L},mathscr{L},leq)$? Can I do this? Is yes, so $underline{mathfrak{B}}(mathscr{L},mathscr{L},leq)$ is not $mathscr{L}$?
For instance. Take a non-complete lattice $(mathbb{N},leq$) and the formal context $(mathbb{N},mathbb{N},leq)$. And about the concept lattice of this context?
Thanks for a help!
formal-languages lattice-orders
asked Dec 1 '18 at 14:19
Na'omiNa'omi
24511
$endgroup$
add a comment |
$begingroup$
I am taking an online course of Introduction to Formal Concept Analysis and I'm trying to understand The Basic Theorem.
Well, if $mathscr{L}$ is a complete lattice, so $mathscr{L}cong underline{mathfrak{B}}(mathscr{L},mathscr{L},leq)$ (concept lattice of context $(mathscr{L},mathscr{L},leq)$).
And we know that:
(*) All concept lattice are complete lattice.
My question is: When we define a formal context $(G,M,I)$, the sets $G,M$ don't have 'conditions'. What happens when I take a non-complete lattice $mathscr{L}$ and define the formal context $(mathscr{L},mathscr{L},leq)$? Can I do this? Is yes, so $underline{mathfrak{B}}(mathscr{L},mathscr{L},leq)$ is not $mathscr{L}$?
For instance. Take a non-complete lattice $(mathbb{N},leq$) and the formal context $(mathbb{N},mathbb{N},leq)$. And about the concept lattice of this context?
Thanks for a help!
formal-languages lattice-orders
asked Dec 1 '18 at 14:19
Na'omiNa'omi
24511
$endgroup$
add a comment |
$begingroup$
I am taking an online course of Introduction to Formal Concept Analysis and I'm trying to understand The Basic Theorem.
Well, if $mathscr{L}$ is a complete lattice, so $mathscr{L}cong underline{mathfrak{B}}(mathscr{L},mathscr{L},leq)$ (concept lattice of context $(mathscr{L},mathscr{L},leq)$).
And we know that:
(*) All concept lattice are complete lattice.
My question is: When we define a formal context $(G,M,I)$, the sets $G,M$ don't have 'conditions'. What happens when I take a non-complete lattice $mathscr{L}$ and define the formal context $(mathscr{L},mathscr{L},leq)$? Can I do this? Is yes, so $underline{mathfrak{B}}(mathscr{L},mathscr{L},leq)$ is not $mathscr{L}$?
For instance. Take a non-complete lattice $(mathbb{N},leq$) and the formal context $(mathbb{N},mathbb{N},leq)$. And about the concept lattice of this context?
Thanks for a help!
formal-languages lattice-orders
asked Dec 1 '18 at 14:19
Na'omiNa'omi
24511
$endgroup$
I am taking an online course of Introduction to Formal Concept Analysis and I'm trying to understand The Basic Theorem.
Well, if $mathscr{L}$ is a complete lattice, so $mathscr{L}cong underline{mathfrak{B}}(mathscr{L},mathscr{L},leq)$ (concept lattice of context $(mathscr{L},mathscr{L},leq)$).
And we know that:
(*) All concept lattice are complete lattice.
My question is: When we define a formal context $(G,M,I)$, the sets $G,M$ don't have 'conditions'. What happens when I take a non-complete lattice $mathscr{L}$ and define the formal context $(mathscr{L},mathscr{L},leq)$? Can I do this? Is yes, so $underline{mathfrak{B}}(mathscr{L},mathscr{L},leq)$ is not $mathscr{L}$?
For instance. Take a non-complete lattice $(mathbb{N},leq$) and the formal context $(mathbb{N},mathbb{N},leq)$. And about the concept lattice of this context?
Thanks for a help!
formal-languages lattice-orders
formal-languages lattice-orders
asked Dec 1 '18 at 14:19
Na'omiNa'omi
24511
asked Dec 1 '18 at 14:19
Na'omiNa'omi
24511
edited Dec 1 '18 at 14:37
Na'omi
asked Dec 1 '18 at 14:19
Na'omiNa'omi
24511
asked Dec 1 '18 at 14:19
Na'omiNa'omi
24511
asked Dec 1 '18 at 14:19
Na'omiNa'omi
24511
24511
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Formal concepts can be identified with certain subsets of $G$ (specifically with the closed subsets in the Galois connection induced by $I$), which are closed under taking arbitrary intersections.
Note that it is happening in the already complete lattice of the power set.
The case $(mathcal L, mathcal L, le)$ with $mathcal L$ a complete lattice is special. Then, for subsets $A, Bsubseteqmathcal L$ we have
$$forall ain A,,bin B, (ale b) iff bigvee Alebigwedge B$$
so that the subsets can be represented by single elements.
answered Dec 1 '18 at 15:25
BerciBerci
60.8k23673
$endgroup$
$begingroup$
Berci, thank you very much for the answer. I don't know if because I don't have much knowing... But I don't understand very well the relation between your answer and my doubt about, for instance, the concept lattice $(mathbb{N},mathbb{N}, leq)$, remembering that $(mathbb{N}, leq)$ is NOT a complete lattice. So, $underline{mathfrak{B}}(mathbb{N},mathbb{N}, leq)$ is a complete lattice and it's NOT similar to $(mathbb{N}, leq)$...? How can I see this...? Thank you very much.
$endgroup$
– Na'omi
Dec 1 '18 at 20:22
1
$begingroup$
Yes, in the case of $Bbb N$, it will be very similar to $Bbb N$, basically only that it's extended by an $infty$ element that makes the lattice complete, namely by the concept $(Bbb N, emptyset)$. All other concepts are of the form $({0,1,dots, n},, {n, n+1,dots})$ for some $n$, which can be identified by $Bbb N$. For a general lattice/poset, the extent is always downward closed, and the intent is upward closed.
$endgroup$
– Berci
Dec 1 '18 at 23:06
$begingroup$
Oh, $(mathbb{N},emptyset)$... So, the professor says: "$(mathbb{N},leq)$ is not a complete lattice because $mathbb{N}$ don't have a supremum"... So, if I understood right, the case is following: When I use sup in definition of complete lattice, I presume that this sup is ON the set (in this case, it would be $mathbb{N}$). However, this don't occur, because the unique sup I could take is $emptyset$ and $emptyset$ isn't an element of $mathbb{N}$. Altough, I can take the concept $(mathbb{N},emptyset)$... It's a little confused, rs... But I thank you very much for the great help!
$endgroup$
– Na'omi
Dec 2 '18 at 14:11
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Formal concepts can be identified with certain subsets of $G$ (specifically with the closed subsets in the Galois connection induced by $I$), which are closed under taking arbitrary intersections.
Note that it is happening in the already complete lattice of the power set.
The case $(mathcal L, mathcal L, le)$ with $mathcal L$ a complete lattice is special. Then, for subsets $A, Bsubseteqmathcal L$ we have
$$forall ain A,,bin B, (ale b) iff bigvee Alebigwedge B$$
so that the subsets can be represented by single elements.
answered Dec 1 '18 at 15:25
BerciBerci
60.8k23673
$endgroup$
$begingroup$
Berci, thank you very much for the answer. I don't know if because I don't have much knowing... But I don't understand very well the relation between your answer and my doubt about, for instance, the concept lattice $(mathbb{N},mathbb{N}, leq)$, remembering that $(mathbb{N}, leq)$ is NOT a complete lattice. So, $underline{mathfrak{B}}(mathbb{N},mathbb{N}, leq)$ is a complete lattice and it's NOT similar to $(mathbb{N}, leq)$...? How can I see this...? Thank you very much.
$endgroup$
– Na'omi
Dec 1 '18 at 20:22
1
$begingroup$
Yes, in the case of $Bbb N$, it will be very similar to $Bbb N$, basically only that it's extended by an $infty$ element that makes the lattice complete, namely by the concept $(Bbb N, emptyset)$. All other concepts are of the form $({0,1,dots, n},, {n, n+1,dots})$ for some $n$, which can be identified by $Bbb N$. For a general lattice/poset, the extent is always downward closed, and the intent is upward closed.
$endgroup$
– Berci
Dec 1 '18 at 23:06
$begingroup$
Oh, $(mathbb{N},emptyset)$... So, the professor says: "$(mathbb{N},leq)$ is not a complete lattice because $mathbb{N}$ don't have a supremum"... So, if I understood right, the case is following: When I use sup in definition of complete lattice, I presume that this sup is ON the set (in this case, it would be $mathbb{N}$). However, this don't occur, because the unique sup I could take is $emptyset$ and $emptyset$ isn't an element of $mathbb{N}$. Altough, I can take the concept $(mathbb{N},emptyset)$... It's a little confused, rs... But I thank you very much for the great help!
$endgroup$
– Na'omi
Dec 2 '18 at 14:11
add a comment |
$begingroup$
Formal concepts can be identified with certain subsets of $G$ (specifically with the closed subsets in the Galois connection induced by $I$), which are closed under taking arbitrary intersections.
Note that it is happening in the already complete lattice of the power set.
The case $(mathcal L, mathcal L, le)$ with $mathcal L$ a complete lattice is special. Then, for subsets $A, Bsubseteqmathcal L$ we have
$$forall ain A,,bin B, (ale b) iff bigvee Alebigwedge B$$
so that the subsets can be represented by single elements.
answered Dec 1 '18 at 15:25
BerciBerci
60.8k23673
$endgroup$
$begingroup$
Berci, thank you very much for the answer. I don't know if because I don't have much knowing... But I don't understand very well the relation between your answer and my doubt about, for instance, the concept lattice $(mathbb{N},mathbb{N}, leq)$, remembering that $(mathbb{N}, leq)$ is NOT a complete lattice. So, $underline{mathfrak{B}}(mathbb{N},mathbb{N}, leq)$ is a complete lattice and it's NOT similar to $(mathbb{N}, leq)$...? How can I see this...? Thank you very much.
$endgroup$
– Na'omi
Dec 1 '18 at 20:22
1
$begingroup$
Yes, in the case of $Bbb N$, it will be very similar to $Bbb N$, basically only that it's extended by an $infty$ element that makes the lattice complete, namely by the concept $(Bbb N, emptyset)$. All other concepts are of the form $({0,1,dots, n},, {n, n+1,dots})$ for some $n$, which can be identified by $Bbb N$. For a general lattice/poset, the extent is always downward closed, and the intent is upward closed.
$endgroup$
– Berci
Dec 1 '18 at 23:06
$begingroup$
Oh, $(mathbb{N},emptyset)$... So, the professor says: "$(mathbb{N},leq)$ is not a complete lattice because $mathbb{N}$ don't have a supremum"... So, if I understood right, the case is following: When I use sup in definition of complete lattice, I presume that this sup is ON the set (in this case, it would be $mathbb{N}$). However, this don't occur, because the unique sup I could take is $emptyset$ and $emptyset$ isn't an element of $mathbb{N}$. Altough, I can take the concept $(mathbb{N},emptyset)$... It's a little confused, rs... But I thank you very much for the great help!
$endgroup$
– Na'omi
Dec 2 '18 at 14:11
add a comment |
$begingroup$
Formal concepts can be identified with certain subsets of $G$ (specifically with the closed subsets in the Galois connection induced by $I$), which are closed under taking arbitrary intersections.
Note that it is happening in the already complete lattice of the power set.
The case $(mathcal L, mathcal L, le)$ with $mathcal L$ a complete lattice is special. Then, for subsets $A, Bsubseteqmathcal L$ we have
$$forall ain A,,bin B, (ale b) iff bigvee Alebigwedge B$$
so that the subsets can be represented by single elements.
answered Dec 1 '18 at 15:25
BerciBerci
60.8k23673
$endgroup$
Formal concepts can be identified with certain subsets of $G$ (specifically with the closed subsets in the Galois connection induced by $I$), which are closed under taking arbitrary intersections.
Note that it is happening in the already complete lattice of the power set.
The case $(mathcal L, mathcal L, le)$ with $mathcal L$ a complete lattice is special. Then, for subsets $A, Bsubseteqmathcal L$ we have
$$forall ain A,,bin B, (ale b) iff bigvee Alebigwedge B$$
so that the subsets can be represented by single elements.
answered Dec 1 '18 at 15:25
BerciBerci
60.8k23673
answered Dec 1 '18 at 15:25
BerciBerci
60.8k23673
answered Dec 1 '18 at 15:25
BerciBerci
60.8k23673
answered Dec 1 '18 at 15:25
BerciBerci
60.8k23673
60.8k23673
$begingroup$
Berci, thank you very much for the answer. I don't know if because I don't have much knowing... But I don't understand very well the relation between your answer and my doubt about, for instance, the concept lattice $(mathbb{N},mathbb{N}, leq)$, remembering that $(mathbb{N}, leq)$ is NOT a complete lattice. So, $underline{mathfrak{B}}(mathbb{N},mathbb{N}, leq)$ is a complete lattice and it's NOT similar to $(mathbb{N}, leq)$...? How can I see this...? Thank you very much.
$endgroup$
– Na'omi
Dec 1 '18 at 20:22
1
$begingroup$
Yes, in the case of $Bbb N$, it will be very similar to $Bbb N$, basically only that it's extended by an $infty$ element that makes the lattice complete, namely by the concept $(Bbb N, emptyset)$. All other concepts are of the form $({0,1,dots, n},, {n, n+1,dots})$ for some $n$, which can be identified by $Bbb N$. For a general lattice/poset, the extent is always downward closed, and the intent is upward closed.
$endgroup$
– Berci
Dec 1 '18 at 23:06
$begingroup$
Oh, $(mathbb{N},emptyset)$... So, the professor says: "$(mathbb{N},leq)$ is not a complete lattice because $mathbb{N}$ don't have a supremum"... So, if I understood right, the case is following: When I use sup in definition of complete lattice, I presume that this sup is ON the set (in this case, it would be $mathbb{N}$). However, this don't occur, because the unique sup I could take is $emptyset$ and $emptyset$ isn't an element of $mathbb{N}$. Altough, I can take the concept $(mathbb{N},emptyset)$... It's a little confused, rs... But I thank you very much for the great help!
$endgroup$
– Na'omi
Dec 2 '18 at 14:11
add a comment |
$begingroup$
Berci, thank you very much for the answer. I don't know if because I don't have much knowing... But I don't understand very well the relation between your answer and my doubt about, for instance, the concept lattice $(mathbb{N},mathbb{N}, leq)$, remembering that $(mathbb{N}, leq)$ is NOT a complete lattice. So, $underline{mathfrak{B}}(mathbb{N},mathbb{N}, leq)$ is a complete lattice and it's NOT similar to $(mathbb{N}, leq)$...? How can I see this...? Thank you very much.
$endgroup$
– Na'omi
Dec 1 '18 at 20:22
1
$begingroup$
Yes, in the case of $Bbb N$, it will be very similar to $Bbb N$, basically only that it's extended by an $infty$ element that makes the lattice complete, namely by the concept $(Bbb N, emptyset)$. All other concepts are of the form $({0,1,dots, n},, {n, n+1,dots})$ for some $n$, which can be identified by $Bbb N$. For a general lattice/poset, the extent is always downward closed, and the intent is upward closed.
$endgroup$
– Berci
Dec 1 '18 at 23:06
$begingroup$
Oh, $(mathbb{N},emptyset)$... So, the professor says: "$(mathbb{N},leq)$ is not a complete lattice because $mathbb{N}$ don't have a supremum"... So, if I understood right, the case is following: When I use sup in definition of complete lattice, I presume that this sup is ON the set (in this case, it would be $mathbb{N}$). However, this don't occur, because the unique sup I could take is $emptyset$ and $emptyset$ isn't an element of $mathbb{N}$. Altough, I can take the concept $(mathbb{N},emptyset)$... It's a little confused, rs... But I thank you very much for the great help!
$endgroup$
– Na'omi
Dec 2 '18 at 14:11
$begingroup$
Berci, thank you very much for the answer. I don't know if because I don't have much knowing... But I don't understand very well the relation between your answer and my doubt about, for instance, the concept lattice $(mathbb{N},mathbb{N}, leq)$, remembering that $(mathbb{N}, leq)$ is NOT a complete lattice. So, $underline{mathfrak{B}}(mathbb{N},mathbb{N}, leq)$ is a complete lattice and it's NOT similar to $(mathbb{N}, leq)$...? How can I see this...? Thank you very much.
$endgroup$
– Na'omi
Dec 1 '18 at 20:22
$begingroup$
Berci, thank you very much for the answer. I don't know if because I don't have much knowing... But I don't understand very well the relation between your answer and my doubt about, for instance, the concept lattice $(mathbb{N},mathbb{N}, leq)$, remembering that $(mathbb{N}, leq)$ is NOT a complete lattice. So, $underline{mathfrak{B}}(mathbb{N},mathbb{N}, leq)$ is a complete lattice and it's NOT similar to $(mathbb{N}, leq)$...? How can I see this...? Thank you very much.
$endgroup$
– Na'omi
Dec 1 '18 at 20:22
1
1
$begingroup$
Yes, in the case of $Bbb N$, it will be very similar to $Bbb N$, basically only that it's extended by an $infty$ element that makes the lattice complete, namely by the concept $(Bbb N, emptyset)$. All other concepts are of the form $({0,1,dots, n},, {n, n+1,dots})$ for some $n$, which can be identified by $Bbb N$. For a general lattice/poset, the extent is always downward closed, and the intent is upward closed.
$endgroup$
– Berci
Dec 1 '18 at 23:06
$begingroup$
Yes, in the case of $Bbb N$, it will be very similar to $Bbb N$, basically only that it's extended by an $infty$ element that makes the lattice complete, namely by the concept $(Bbb N, emptyset)$. All other concepts are of the form $({0,1,dots, n},, {n, n+1,dots})$ for some $n$, which can be identified by $Bbb N$. For a general lattice/poset, the extent is always downward closed, and the intent is upward closed.
$endgroup$
– Berci
Dec 1 '18 at 23:06
$begingroup$
Oh, $(mathbb{N},emptyset)$... So, the professor says: "$(mathbb{N},leq)$ is not a complete lattice because $mathbb{N}$ don't have a supremum"... So, if I understood right, the case is following: When I use sup in definition of complete lattice, I presume that this sup is ON the set (in this case, it would be $mathbb{N}$). However, this don't occur, because the unique sup I could take is $emptyset$ and $emptyset$ isn't an element of $mathbb{N}$. Altough, I can take the concept $(mathbb{N},emptyset)$... It's a little confused, rs... But I thank you very much for the great help!
$endgroup$
– Na'omi
Dec 2 '18 at 14:11
$begingroup$
Oh, $(mathbb{N},emptyset)$... So, the professor says: "$(mathbb{N},leq)$ is not a complete lattice because $mathbb{N}$ don't have a supremum"... So, if I understood right, the case is following: When I use sup in definition of complete lattice, I presume that this sup is ON the set (in this case, it would be $mathbb{N}$). However, this don't occur, because the unique sup I could take is $emptyset$ and $emptyset$ isn't an element of $mathbb{N}$. Altough, I can take the concept $(mathbb{N},emptyset)$... It's a little confused, rs... But I thank you very much for the great help!
$endgroup$
– Na'omi
Dec 2 '18 at 14:11
add a comment |
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