Determine if the structure is a lattice

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Given $X =$ $1, 2, 3, 4$ and $S = $ 1, 2, 2, ∅, 3, 4, X the
ordered set through the relation of inclusion $⊂$.



(a) Draw the corresponding Hasse diagram of $S$.



(b) Determine if $S$ is a lattice.




The hasse diagaram would start from the $∅$ element then connected to $1,2,3,4$
all connecting together to make $X$ with inter-connections between them if I am not wrong. But how to determine the lattice exactly?




abstract-algebra group-theory lattice-orders




share|cite|improve this question



















  • Look at the axioms for a lattice and check them. Do all the necessary upper bounds exist?
    – Ethan Bolker
    Jul 16 at 15:55










  • I think the upper bound will be unique and it will be $X$ if I am not wrong right?
    – CptPackage
    Jul 16 at 15:58










  • If you look at the inclusion connections between the sets in $S$, you'll see that the Hasse diagram is that of the pentagon, $mathbf N_5$, as in this question, with $C=3,4$, $B=2$ and $A=1,2$. Of course I'm supposing that $X=1,2,3,4$ or a bigger set.
    – amrsa
    Jul 16 at 17:19










  • @amrsa Shouldn't it be more like this?
    – CptPackage
    Jul 16 at 19:15











  • How could it be like that if $|S|=5$? The diagram you have in that link has 8 nodes.
    – amrsa
    Jul 16 at 21:57














up vote
0
down vote

favorite
1













Given $X =$ $1, 2, 3, 4$ and $S = $ 1, 2, 2, ∅, 3, 4, X the
ordered set through the relation of inclusion $⊂$.



(a) Draw the corresponding Hasse diagram of $S$.



(b) Determine if $S$ is a lattice.




The hasse diagaram would start from the $∅$ element then connected to $1,2,3,4$
all connecting together to make $X$ with inter-connections between them if I am not wrong. But how to determine the lattice exactly?




abstract-algebra group-theory lattice-orders




share|cite|improve this question



















  • Look at the axioms for a lattice and check them. Do all the necessary upper bounds exist?
    – Ethan Bolker
    Jul 16 at 15:55










  • I think the upper bound will be unique and it will be $X$ if I am not wrong right?
    – CptPackage
    Jul 16 at 15:58










  • If you look at the inclusion connections between the sets in $S$, you'll see that the Hasse diagram is that of the pentagon, $mathbf N_5$, as in this question, with $C=3,4$, $B=2$ and $A=1,2$. Of course I'm supposing that $X=1,2,3,4$ or a bigger set.
    – amrsa
    Jul 16 at 17:19










  • @amrsa Shouldn't it be more like this?
    – CptPackage
    Jul 16 at 19:15











  • How could it be like that if $|S|=5$? The diagram you have in that link has 8 nodes.
    – amrsa
    Jul 16 at 21:57












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1






Given $X =$ $1, 2, 3, 4$ and $S = $ 1, 2, 2, ∅, 3, 4, X the
ordered set through the relation of inclusion $⊂$.



(a) Draw the corresponding Hasse diagram of $S$.



(b) Determine if $S$ is a lattice.




The hasse diagaram would start from the $∅$ element then connected to $1,2,3,4$
all connecting together to make $X$ with inter-connections between them if I am not wrong. But how to determine the lattice exactly?




abstract-algebra group-theory lattice-orders




share|cite|improve this question












Given $X =$ $1, 2, 3, 4$ and $S = $ 1, 2, 2, ∅, 3, 4, X the
ordered set through the relation of inclusion $⊂$.



(a) Draw the corresponding Hasse diagram of $S$.



(b) Determine if $S$ is a lattice.




The hasse diagaram would start from the $∅$ element then connected to $1,2,3,4$
all connecting together to make $X$ with inter-connections between them if I am not wrong. But how to determine the lattice exactly?





abstract-algebra group-theory lattice-orders





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 16 at 15:53









CptPackage

417




417











  • Look at the axioms for a lattice and check them. Do all the necessary upper bounds exist?
    – Ethan Bolker
    Jul 16 at 15:55










  • I think the upper bound will be unique and it will be $X$ if I am not wrong right?
    – CptPackage
    Jul 16 at 15:58










  • If you look at the inclusion connections between the sets in $S$, you'll see that the Hasse diagram is that of the pentagon, $mathbf N_5$, as in this question, with $C=3,4$, $B=2$ and $A=1,2$. Of course I'm supposing that $X=1,2,3,4$ or a bigger set.
    – amrsa
    Jul 16 at 17:19










  • @amrsa Shouldn't it be more like this?
    – CptPackage
    Jul 16 at 19:15











  • How could it be like that if $|S|=5$? The diagram you have in that link has 8 nodes.
    – amrsa
    Jul 16 at 21:57
















  • Look at the axioms for a lattice and check them. Do all the necessary upper bounds exist?
    – Ethan Bolker
    Jul 16 at 15:55










  • I think the upper bound will be unique and it will be $X$ if I am not wrong right?
    – CptPackage
    Jul 16 at 15:58










  • If you look at the inclusion connections between the sets in $S$, you'll see that the Hasse diagram is that of the pentagon, $mathbf N_5$, as in this question, with $C=3,4$, $B=2$ and $A=1,2$. Of course I'm supposing that $X=1,2,3,4$ or a bigger set.
    – amrsa
    Jul 16 at 17:19










  • @amrsa Shouldn't it be more like this?
    – CptPackage
    Jul 16 at 19:15











  • How could it be like that if $|S|=5$? The diagram you have in that link has 8 nodes.
    – amrsa
    Jul 16 at 21:57















Look at the axioms for a lattice and check them. Do all the necessary upper bounds exist?
– Ethan Bolker
Jul 16 at 15:55




Look at the axioms for a lattice and check them. Do all the necessary upper bounds exist?
– Ethan Bolker
Jul 16 at 15:55












I think the upper bound will be unique and it will be $X$ if I am not wrong right?
– CptPackage
Jul 16 at 15:58




I think the upper bound will be unique and it will be $X$ if I am not wrong right?
– CptPackage
Jul 16 at 15:58












If you look at the inclusion connections between the sets in $S$, you'll see that the Hasse diagram is that of the pentagon, $mathbf N_5$, as in this question, with $C=3,4$, $B=2$ and $A=1,2$. Of course I'm supposing that $X=1,2,3,4$ or a bigger set.
– amrsa
Jul 16 at 17:19




If you look at the inclusion connections between the sets in $S$, you'll see that the Hasse diagram is that of the pentagon, $mathbf N_5$, as in this question, with $C=3,4$, $B=2$ and $A=1,2$. Of course I'm supposing that $X=1,2,3,4$ or a bigger set.
– amrsa
Jul 16 at 17:19












@amrsa Shouldn't it be more like this?
– CptPackage
Jul 16 at 19:15





@amrsa Shouldn't it be more like this?
– CptPackage
Jul 16 at 19:15













How could it be like that if $|S|=5$? The diagram you have in that link has 8 nodes.
– amrsa
Jul 16 at 21:57




How could it be like that if $|S|=5$? The diagram you have in that link has 8 nodes.
– amrsa
Jul 16 at 21:57















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