I'm studying from Michael Carter's "Foundations", and on page 29 he makes the comment, "Note that the requirement of being a sublattice is more stringent than being a complete lattice in its own right".

This seems counterintuitive. Some lattice $L$ is complete if every nonempty subset $S\subseteq L$ has a least upper bound and greatest lower bound in $L$. This is a more stringent requirement than merely needing bounds for every combination of two points in $L$. By the same token, requiring that all subsets $T \subseteq S$ have meets and joins in $S$ (thereby making $S$ complete) ought to be more stringent than merely requiring $S$ to be a lattice, right? In what sense are the requirements for being a sublattice more stringent on $S$?

One more related question – suppose $X= \{ (1,1),(2,1),(3,1),(1,2),(1,3),(3,3) \}$. The book says $X$ is a complete lattice, but not a sublattice of $\{1,2,3\}\times \{1,2,3\}$. The point $(3,3)$ is an upper bound for all pairs of points in $X$, so shouldn't that be enough to qualify $X$ as a sublattice of the set above? Why is $X$ not considered a sublattice?

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    $\begingroup$ Take $\mathcal{L} = [0,2]$ with the usual total order on the reals, which is a complete lattice. Let $L=[0,1)\cup\{2\}$; this is a lattice under the usual order of the reals, and as a lattice it is a sublattice of $\mathcal{L}$; it is also a complete lattice, but it is not a sublattice of $\mathcal{L}$ because the least upper bound of $[0,1)$ in $\mathcal{L}$ does not agree with the least upper bound in $L$. $\endgroup$ May 16, 2011 at 0:36
  • $\begingroup$ @ArturoMagidin I don't get from this comment if L is a sublattice or not of [0,2]? You first say : "...is a lattice under the usual order of the reals, and as a lattice it is a sublattice of [0,2]" and then " ..it is not a sublattice because....". $\endgroup$ Jan 31, 2019 at 12:40
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    $\begingroup$ @GabrieleScarlatti: Lattices only require finite joins and meets; so $L$ is a sublattice (it is a lattice under the induced operations); it is not a complete sublattice because it does not respect the infinite joins and meets: the join of all of $[0,1)$ in $L$ is different from the joint of $[0,1]$ in $\mathcal{L}$. $\endgroup$ Jan 31, 2019 at 17:03

1 Answer 1


The meet and join operations in a sublattice must agree with the meet and join operations in the lattice. Consider your example. The least upper bound, in $X$, of $(1,2)$ and $(2,1)$ is $(3,3)$. In the space $\{ 1,2,3 \} \times \{ 1, 2, 3 \} $ the least upper bound of $(1,2)$ and $(2,1)$ is $(2,2)$.

  • $\begingroup$ Oh wow, that's a pretty big difference. Thanks for the help. $\endgroup$ May 16, 2011 at 0:36
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    $\begingroup$ This is a very common point of confusion: there is a difference between being a sublattice and being a subposet that happens to be a lattice. The same problem comes up in the context of subtrees. $\endgroup$ May 16, 2011 at 0:36

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