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This question already has an answer here:

I know that a chain is always a lattice (https://math.stackexchange.com/questions/974596/is-every-chain-a-lattice#=), where

  1. A chain is a subset of a poset in which every two elements are comparable and
  2. A lattice is a poset in which every finite subset has a greatest lower bound and a lowest upper bound.

What would be an example of a lattice that isn't a chain? Or are they equivalent?

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marked as duplicate by Alex Provost, Community Jul 22 '17 at 17:02

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  • $\begingroup$ $(P(X), \subset)$ where $X$ is any set with at least two elements, and where $P(X)$ denotes the power set. This is a complete lattice that isn't a chain $\endgroup$ – Max Jul 22 '17 at 15:22
  • $\begingroup$ Subgroups of groups also provide a range of examples - the non-cyclic group of order 4, or the groups of order 6 are simple ones. $\endgroup$ – Mark Bennet Jul 22 '17 at 15:42
  • $\begingroup$ The empty subset is finite. Does your definition of lattice include a lub, glb for the empty set? That would mean a lattice would have to have a bottom and top element. Is that true? Are the reals with the usual order a lattice or not? $\endgroup$ – William Elliot Jul 22 '17 at 20:45
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There are plenty of examples: a very standard one:

Let $X$ be a set of at least 2 elements and let the lattice be the set $\mathscr{P}(X)$ (its powerset (i.e. all its subsets)), ordered by inclusion. This is a lattice with $\inf(A,B) = A \land B = A \cap B$ and $\sup(A,B) = A \lor B = A \cup B$. The elements $\{a\}$ and $\{b\}$ are incomparable for $a,b \in X, a \neq b$

Or take any positive integer $N$ with at least two prime factors and take the set of all divisors of $N$, ordered by divisibility: $a \le b$ iff $a$ divides $b$ evenly. $\sup(a,b) = \operatorname{lcm}(a,b)$ and $\inf(a,b) = \operatorname{gcd}(a,b)$; different prime divisors are incomparable.

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