In the proof of Theorem 2, you write “Clearly mathcal{A} 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 mathcal{A} would be as well.
But
) is not convex. Project
) onto
in the coordinates corresponding to the sentences phi 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!
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 f is closed. Let Gammasubsetmathcal{A}timesmathcal{A} be the graph of f. What you check is that, if
) is a sequence of points in Gamma which approaches a limit, then that limit is in Gamma. This set off alarm bells in my head, because there are examples of a topological space X, and a subspace GammasubsetX, so that Gamma is not closed in X but, if x_i is any sequence in Gamma which approaches a limit in X, then that limit is in Gamma. See Wikipedia’s article on sequential spaces. However, this is not an actual problem. Since L′ is countable,
is metrizable and therefore closure is the same as sequential closure in
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.
In the proof of Theorem 2, you write “Clearly mathcal{A} is convex.” This isn’t clear to me; could you explain what I am missing?
More specifically, let
But
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!
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 f is closed. Let Gammasubsetmathcal{A}timesmathcal{A} be the graph of f. What you check is that, if
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.