# Example of a bounded lattice that is NOT complete

I know that every complete lattice is bounded. Is there a simple example for a bounded lattice that is not complete?

Thank you

• What are your thoughts on this? :)
– Shaun
Commented May 18, 2014 at 15:25
• I don't really have a clue. I thought that the answer to this question would help me to get a better understanding of the lattice theory Commented May 18, 2014 at 16:00

I guess the simplest example is $$\{\frac1n:n\in\mathbb N\}\cup\{-\frac1n:n\in\mathbb N\}$$ with the usual ordering of real numbers. Perhaps the most natural example is $$\{X\subseteq\mathbb N:X\text{ is finite or }\mathbb N\setminus X\text{ is finite }\}$$ ordered by set inclusion. In the first example the set of all negative numbers, in the second example the set of all finite sets of even numbers has no least upper bound.

• Your first example is not bounded, so it is not what OP wanted. In your second example, $\mathbb N$ is the least upper bound of the set of all finite sets. One has to take (for example) $$A=\{Y\subseteq{\mathbb N}:Y\text{ is finite and contains only even numbers}\}.$$ The set A has no least upper bound. Commented May 19, 2014 at 11:31
• Heh, forget about what I said about the first example. My apologies. Commented May 19, 2014 at 11:42
• @GejzaJenča Thanks for your correction to my second example.
– bof
Commented May 19, 2014 at 11:43

Clearly, $\mathbb{Q}\cap[0,1]$ is a bounded lattice. Pick an irational number $x\in[0,1]$ and observe that $\mathbb{Q}\cap [0,x)$ has no join.

• How could you pick x as an irrational number when the lattice was $\mathbb{Q}\cap [0,1]$?
– Shub
Commented May 4 at 11:30
• @Shub x does not need to belong to the lattice. It is used just to specify the subset that lacks a join. Commented May 5 at 10:03

Update: My answer below is wrong! (Thanks to bof for pointing that out.) I will leave the answer here because I think my mistake and bof's comment could maybe be instructive.

Let $\mathbb{N}$ be the set of natural numbers. Let $\mathcal{P}_{fin}(\mathbb{N})$ denote the collection of finite subsets of $\mathbb{N}$. Then $L=\mathcal{P}_{fin}(\mathbb{N})\cup\{\mathbb{N}\}$ is a bounded lattice under inclusion. However, it is not complete since $\{0\}\cup\{2\}\cup\{4\}\cup\{6\}\cup\ldots$ is not in $L$.

• Actually, your example $L$ is a complete lattice: every subset of $L$ has a greatest lower bound, namely, its set-theoretic intersection; therefore, every subset of $L$ has a least upper bound as well. If a set $X\subseteq L$ has no finite upper bound, then its least (and only) upper bound is $\mathbb N$; in particular,$$\sup\{\{0\},\{2\},\{4\},\dots\}=\mathbb N.$$
– bof
Commented May 18, 2014 at 21:04