Benja comments on Reflection in Probabilistic Logic - Less Wrong

63 Post author: Eliezer_Yudkowsky 24 March 2013 04:37PM

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Comment author: Benja 23 March 2013 06:52:34PM * 8 points [-]

Comments on the proof of Theorem 2:

I believe the entire second-to-last paragraph can be replaced by:

If there is no such $\varphi \in A_{\mathbb{P}_1}$ , then $\mathbb{P}_2\notin\mathcal{A}$ . Thus we can take $U_2$ to be the complement of $\mathcal{A}$ (which is open since $\mathcal{A}$ is closed) and let $U_1$ be the entire space.

I'm thinking that the proof becomes conceptually slightly clearer if you show that the graph is closed by showing that it contains the limit of any convergent sequence, though:

Suppose $\mathbb{P}_n\in\mathcal{A}(\mathbb{P}'_n)$ for all $n\in\mathbb{N}$ , and $(\mathbb{P}'_n,\mathbb{P}_n)\to(\mathbb{P}',\mathbb{P})$ . Since $\mathcal{A}$ is closed, we have $\mathbb{P}\in\mathcal{A}$ . We must show that $\mathbb{P}(A_{\mathbb{P}'}) = 1$ . Thus, suppose that $a<\mathbb{P}'(\varphi)<b$ . Since $\mathbb{P}'_n(\varphi)\to\mathbb{P}(\varphi)$ , for all sufficiently large $n$ we have $a<\mathbb{P}'_n(\varphi)<b$ and therefore $\mathbb{P}_n(a < \mathbb{P}(\ulcorner\varphi\urcorner) < b) = 1$ . Since $\mathbb{P}_n\to\mathbb{P}$ , it follows that $\mathbb{P}(a < \mathbb{P}(\ulcorner\varphi\urcorner) < b) = 1$ . In other words, $\mathbb{P}$ assigns probability 1 to every element of $A_{\mathbb{P}'}$ , and hence to all of $A_{\mathbb{P}'}$ (since this set is countable).