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A naive question.

Let $S$ be a set and let $[0,1]^S$ the set of functions from $S$ to the closed interval $[0,1]$.

Suppose given some function $P \colon [0,1]^S \to [0,1]$ satisfying the following three conditions:

  1. If $f \geq g$ everywhere on $S$, then $P(f) \geq P(g)$;
  2. $P(\min(f,g)) \geq \min(P(f),P(g))$;
  3. $P(1-f) = 1 - P(f)$.

This is supposed to model a situation each point in $S$ has a "degree of belief" in some proposition, which yields a function $f$ in $[0,1]^S$; then $P$ is a process which takes all these degrees of belief and aggregates them into a "consensus" degree of belief $P(f)$.

Of course, this is meant to mimic the definition of an ultrafilter, which I think is given by the above definition with [0,1] replaced by {0,1}.

Certainly you have "principal" $P$, which just evaluate $f$ at some point $s$ of $S$. I suppose you could get other $P$ by sending $f$ to its limit with respect to some non-principal ultrafilter.

Is that it?

Added: Actually, the second condition above is perhaps too strong. I don't see an option for "hide question until I've thought about a bit more about what the best version of the question is" so I will just append this remark.

Added: Thanks, guys, for all the great answers. I now think the formulation of (2) was misguided (at least if the definition is meant to model consensus about degrees of belief) and I don't know what the "right" formulation is. One might well, for instance, want P to behave well when f and g refer to independent propositions; that would ask that P(fg) = P(f)P(g), which in the case of {0,1}-valued functions again agrees with the ultrafilter definition. This rules out averages but leaves in evaluation at ultrafilters.

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Could it be that the inequality in (2) should be $\leq$? –  Kevin O'Bryant Mar 7 '11 at 4:00
    
Oh, I've missed the boat; that actually follows from (1), right? Because min(f,g) is below both f and g everywhere. So do I actually want an EQUALITY there? –  JSE Mar 7 '11 at 4:13
    
Perhaps you also want $P$ to send the constant function $s\mapsto c$ to $c$? Otherwise you also get things like the limit of $g\circ f$ with respect to some ultrafilter, where $g: [0,1]\to [0,1]$ is monotone and symmetric. –  Noah Stein Mar 7 '11 at 4:24
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I haven't thought much about this, but your $P$ reminds me of integration with respect to a finitely additive measure. –  Nate Eldredge Mar 7 '11 at 13:13
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Nate, there is a one-to-one correspondence between finitely additive probability measures on $S$ and states on $\ell^{\infty}(S,\mathbb R$); see my answer below. –  Andreas Thom Mar 7 '11 at 13:24
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2 Answers

A natural class of functions $P: [0,1]^S \to [0,1]$ which generalizes ultrafilters in a probabilistic sense is coming from states $\phi \colon \ell^{\infty}(S,\mathbb R) \to \mathbb R$, i.e. linear and monotone functionals with $\phi(1_S)=1$. Each state clearly yields a probabilistic ultrafilter by restriction.

  • Ultrafilters correspond to special such functionals which in addition satisfy $\phi(fg)=\phi(f)\phi(g)$ since $\ell^{\infty}(S)=C(\beta S)$.
  • Any such $\phi$ corresponds to a Radon probability measure on the compact Hausdorff space $\beta S$ by the Riesz representation theorem.
  • Many examples have been studied: Let $S=\mathbb N$, $\omega \in \beta \mathbb N \setminus \mathbb N$ and consider $$\phi(f)=P(f) := \lim_{n \to \omega} \frac1n \sum_{i=1}^n f(i).$$

This example does not come from taking a limit of the function $f$ with respect to some non-principal ultrafilter. In this sense the answer to your question is no.

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The specific example here doesn't satisfy condition 2 of the original question. Let $f(n)$ be 1 for odd $n$ and 0 for even $n$. Then $P(f)=P(1-f)=1/2$ but `$P(\min\{f,1-f\})=P(0)=0$. –  Andreas Blass Mar 7 '11 at 15:34
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Assume that $P$ satisfies the three conditions in the original question plus Noah Stein's additional requirement (declared " very wise" by the OP) that $P(s\mapsto c)=c$ for all constants $c$. I claim that then $P$ is the operation of limit along a certain ultrafilter $U$ on $S$.

As a first step, consider any function $f$ on $S$ that takes only the values 0 and 1; so $f$ is the characteristic function of a subset $A$ of $S$, while $1-f$ is the characteristic function of $S-A$. Conditions 1 and 2 in the question imply that $P$ commutes with min, i.e., that equality holds in condition 2. In particular, since $\min\{f,1-f\}=0$, we have that one of $P(f)$ and $P(1-f)$ is 0, and then, by condition 3, the other one is 1. In particular, the characteristic function of any $A\subseteq S$ is sent by $P$ to 0 or to 1. Then the conditions in the question immediately imply that $U=\{A\subseteq S:P(A)=1\}$ is an ultrafilter on $S$. Clearly, on functions that take only the values 0 and 1, $P$ agrees with limit along $U$; it remains to prove the same for arbitrary functions $f:S\to[0,1]$.

Let $f$ be such a function, and let $l$ be its limit along $U$. Consider an arbitrary $c<l$ and note that, by definition of limit, the set $A=\{s\in S:f(s)>c\}$ is in $U$. Let $g$ be the characteristic function of $A$, and notice that $f\geq\min\{s\mapsto c, g\}$. Since $P$ sends $s\mapsto c$ to $c$ and sends $g$ to 1, it must send $f$ to a value $\geq c$. Similar reasoning (using condition 3 to "turn the picture upside down") shows that $P(f)\leq d$ for all $d>l$. Therefore $P(f)=l$, as required.

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