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DavidS comments on Reflection in Probabilistic Logic - Less Wrong
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In the proof of Theorem 2, you write "Clearly
is convex." This isn't clear to me; could you explain what I am missing?
More specifically, let)
be the subset of 
obeying
%20%3C%20b%20\%20\implies%20\%20\mathbb{P}\left(%20a%20%3C%20\mathbb{P}(\lceil%20\phi%20\rceil)%20%3C%20b%20\right)%20=1%20)
. So }%20X(\phi,a,b))
. If )
were convex, then 
would be as well.
But)
is not convex. Project )
onto 
in the coordinates corresponding to the sentences 
and %20%3C%20b)
. The image is %20\cup%20\left(%20%20[0,1]%20\times%20\{%201%20\}%20\right)%20\cup%20\left(%20[b,1]%20\times%20[0,1]%20\right))
. This is not convex.
Of course, the fact that the $X(\phi,a,b)$ are not convex doesn't mean that their intersection isn't, but it makes it non-obvious to me why the intersection should be convex. Thanks!
The set A is convex because the convex combination (t times one plus (1-t) times the other) of two coherent probability distributions remains a coherent probability distribution. This in turn is because the convex combination of two probability measures over a space of models (cf. definition 1) remains a probability distribution over the space of models.
I think, but am not sure, that your issue is looking at arbitrary points of [0,1]^{L'}, rather than the ones which correspond to probability measures.
Other nitpicks (which I don't think are real problems):
If the Wikipedia article on Kakatuni's fixed point theorem is to be believed, then Kakatuni's result is only for finite dimensional vector spaces. You probably want to be citing either Glicksberg or Fan for the infinite dimensional version. These each have some additional hypotheses, so you should check the additional hypotheses.
At the end of the proof of Theorem 2, you want to check that the graph of
is closed. Let 
be the graph of 
. What you check is that, if )
is a sequence of points in 
which approaches a limit, then that limit is in 
. This set off alarm bells in my head, because there are examples of a topological space 
, and a subspace 
, so that 
is not closed in 
but, if 
is any sequence in 
which approaches a limit in 
, then that limit is in 
. See Wikipedia's article on sequential spaces. However, this is not an actual problem. Since 
is countable, 
is metrizable and therefore closure is the same as sequential closure in 
.