Let us take the set $F$ of all functions from $[0,1]$ to reals of the form: $$f(x) = \sum_{n \in N} q_n x^n$$ such that:
- $0 \le q_n \le 1$ for all naturals $n$
- $\sum_{n \in N} q_n$ = 1
- the set $\{n | q_n >0\}$ is at most countable
Let us define the ordering relations $\sqsubseteq$ over $F$ such that $f \sqsubseteq f'$ if and only if $f(x) \le f'(x)$ for all $x \in [0,1]$.
My conjecture is that this is a complete lattice, but I was unable to prove this. Can anybody help me? Thanks in advance.
The set $F$ you define is not a complete lattice.
Consider the set $A = \{1,x,x^2,x^3,\dots\}$. I claim that $A$ does not have an infimum in $F$. For suppose $f(x) = \sum_n q_n x^n$ is such infimum.
For any $m$ you have $f(x) \leq x^m$ for all $x$. In particular, for any $n,m$ you have $q_nx^n \leq x^m$, and if you take $m > n$ and $x$ small enough, you find that $q_n$ has to be $0$. But this is true for all $n$, so $f(x) = 0$, identically. This contradicts the condition $\sum_n q_n = 1$.
(I'm not sure what happens if you drop the assumption that $\sum_n q_n = 1$. Perhaps this leads to a complete lattice.)