Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $D$ be a set, $\mathbb{N_0}$ the set of natural numbers including zero. Let $P$ be the set of all functions from $D$ to $\mathbb{N_0}$, i.e. $P = \lbrace m \mid m: D \rightarrow \mathbb{N_0} \rbrace$.

Let $f, g \in P$. Then we define that $f \leq g$ holds iff $\forall d\in D f(d) \leq g(d)$. So $\leq$ is a partial order on $P$.

Now, $(P,\leq)$ is a lattice but it is not a complete lattice because some subsets of $P$ don't have a supremum in $P$.

So let us allow the functions to map to an infinity, $\infty$, too, i.e. $\forall n\in \mathbb{N_0} n < \infty$. Let $P' = \lbrace m \mid m: D \rightarrow \mathbb{N_0} \cup \infty \rbrace$.

Is $(P',\leq)$ a complete lattice? Resp., I am quite sure it is and I would like to know if there is some reference I could cite or if it is so trivial that I can simply state it.

share|improve this question
    
This is the kind of reference I was looking for: Discrete t-norms and operations on extended multisets by: J. Casasnovas, G. Mayor Fuzzy Sets and Systems 159, 1165 (2008) dx.doi.org/10.1016/j.fss.2007.12.005 Just in case someone else needs it. –  fraktalek Oct 8 '10 at 6:53

1 Answer 1

up vote 2 down vote accepted

Once one adds infinity, it is easy to check that $(P', \leq)$ is indeed a complete lattice. For any subset $A$ of $P'$, it is easy to see that $\vee A$ is just the pointwise supremum of members of $A$. In the case that $A$ is empty, we have that $\vee A$ is the zero function. So, $(P', \leq)$ is a complete join semi-lattice, and hence a complete lattice.

share|improve this answer
    
Yes, I thought so, I guess I will simply give this argument. Thanks! –  fraktalek Oct 7 '10 at 8:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.