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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.

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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) Just in case someone else needs it. – fraktalek Oct 8 '10 at 6:53
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.

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

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