I was studying about Complete Lattices from the text book J.P. Tremblay. The definition of a complete lattice is that each of its non empty subsets should have a least upper bound and greatest lower bound. By this definition every finite lattice should be complete. But what about Infinite Lattices? Is there an Infinite Lattice which is not Complete. So far I couldn't come up with an example or prove that there are no such case.
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3 Answers
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The set $\mathbb Q$ of all rational numbers, with the usual linear order, is an infinite distributive lattice which is not complete. For example, $\mathbb Q$ itself has neither a least upper bound nor a greatest lower bound in $\mathbb Q$; neither does the bounded nonempty set $\{x\in\mathbb Q:x^2\lt2\}.$
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Consider the lattice of finite subsets of $\Bbb N$. Since the union and intersection of two finite subsets is again finite, this is indeed a lattice. But the set $A=\{\{0\},\{0,1\},\{0,1,2\},\dots\}$ has no upper bound in the lattice. Of course $\Bbb N$ is an upper bound of $A$ in $\mathcal P(\Bbb N)$, the powerset of $\Bbb N$.
answered Oct 28, 2016 at 7:19
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$\begingroup$ This set is also a very important example in set theory that pertains to some theorems about infinite sets either having an infinite totally unordered subset or containing an infinite chain (linear order). $\endgroup$ Oct 28, 2016 at 7:46
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The partially ordered set $(\mathbb R, \leqslant)$ of real numbers is a lattice, but not a complete lattice. Note that the subset $S = (0,\infty)$ does not have a supremum in $\mathbb R$.
answered Mar 31, 2022 at 17:40