Knaster-Tarski to show the existence of a unique fixpoint Let $\mbox{CDF}\subseteq [0,1]^{[0,\infty)}$ be the collection of cummulative distribution functions and let $T:\mbox{CDF} \rightarrow \mbox{CDF}$ be a function. We define the order $F \leq G \Leftrightarrow \forall s \in [0,\infty): F(s) \leq G(s)$. Assume we have $F \leq G \Rightarrow TF \leq TG$, I would then like to show, using Knaster-Tarski, that $T$ has a unique fix point.
I am especially interested in which extra conditions need to be imposed on $T$ to ensure uniqueness?
 A: We can see that this set is a partially ordered set, and not a totally ordered set. The first three properties are straightforward to show, to show that it does not satisfy the fourth property, define $X_1 \sim \textrm{N}(0,.01)$ and 
$X_2 \sim \textrm{N}(0,10)$.
We can see that at first $X_2$ has a higher CDF for $c<<0$, but because of the spikiness of $X_1$ there exists a period after $c=0$ where $X_1$ has a higher CDF.
We need to show that the partial ordered set is a Complete Lattice to satisfy the requirements of the Knaster-Tarski Theorem. 
Suppose we have a subset of CDF's $\mathcal{C}$. We can partition $[0,\infty)$ into $[0,\infty) = \cup_{i}A_i$ where each $A_i$ is a set on which one of the $C_i \in\mathcal{C}$ has the maximal CDF value $\forall a \in A_i$. Define $C^*[a] = C_i[a]$ for $i$ such that $a \in A_i$. It is straightforward to see that $C^*$ is a least upper bound for $\mathcal{C}$. Thus supremum's always exist for all subsets.
To show infimum exist for all sets do an analogous step, but with the minimum CDF on each set $A_i$.
Thus the set of CDFs is a complete lattice, which in conjunction with the order preserving property of the operator $T$ implies that the Knaster-Tarski fixed point theorem.
To use the order preserving property, we can find at least one fixed point in $F[c] = \delta_0(c)$. Because this is the maximal element of the whole space of CDFs
