Can a lattice be bounded and not complete and if yes can you please show an example?
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1See https://en.wikipedia.org/wiki/Lattice_(order)#Bounded_lattice and https://en.wikipedia.org/wiki/Complete_lattice – Angel Nov 19 '21 at 16:17
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1Does this answer your question? Example of a bounded lattice that is NOT complete – hardmath Nov 19 '21 at 18:11
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Consider $\mathbb{Q} \cup \{\pm \infty\}$, which is a total order and hence a bounded lattice. However, there is no supremum of $\{x \in \mathbb{Q} \mid x^2 < 2\}$.
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