Complete Lattice and Concept Lattice











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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!










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    up vote
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    down vote

    favorite
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    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!










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite
      1









      up vote
      1
      down vote

      favorite
      1






      1





      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!










      share|cite|improve this question















      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






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      edited Dec 1 at 14:37

























      asked Dec 1 at 14:19









      Na'omi

      24010




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          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.






          share|cite|improve this answer





















          • 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.
            – Na'omi
            Dec 1 at 20:22








          • 1




            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.
            – Berci
            Dec 1 at 23:06












          • 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!
            – Na'omi
            2 days ago













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          up vote
          1
          down vote



          accepted










          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.






          share|cite|improve this answer





















          • 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.
            – Na'omi
            Dec 1 at 20:22








          • 1




            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.
            – Berci
            Dec 1 at 23:06












          • 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!
            – Na'omi
            2 days ago

















          up vote
          1
          down vote



          accepted










          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.






          share|cite|improve this answer





















          • 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.
            – Na'omi
            Dec 1 at 20:22








          • 1




            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.
            – Berci
            Dec 1 at 23:06












          • 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!
            – Na'omi
            2 days ago















          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 1 at 15:25









          Berci

          59.2k23671




          59.2k23671












          • 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.
            – Na'omi
            Dec 1 at 20:22








          • 1




            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.
            – Berci
            Dec 1 at 23:06












          • 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!
            – Na'omi
            2 days ago




















          • 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.
            – Na'omi
            Dec 1 at 20:22








          • 1




            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.
            – Berci
            Dec 1 at 23:06












          • 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!
            – Na'omi
            2 days ago


















          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.
          – Na'omi
          Dec 1 at 20:22






          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.
          – Na'omi
          Dec 1 at 20:22






          1




          1




          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.
          – Berci
          Dec 1 at 23:06






          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.
          – Berci
          Dec 1 at 23:06














          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!
          – Na'omi
          2 days ago






          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!
          – Na'omi
          2 days ago




















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