User:IssaRice/Logical inductor construction: Difference between revisions
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"Observe that <math>\mathcal V'</math> is equal to the natural inclusion of the finite-dimensional cube <math>[0,1]^{\mathcal S'}</math> in the space of all valuations <math>\mathcal V = [0,1]^{\mathcal S}</math>." -- I think what this is saying is that since <math>\mathcal S' \subset \mathcal S</math>, we can think of <math>[0,1]^{\mathcal S'}</math> as being sort of a subset of <math>[0,1]^{\mathcal S}</math>. Except it's not strictly speaking a subset, since the functions in <math>[0,1]^{\mathcal S'}</math> and <math>[0,1]^{\mathcal S}</math> have different domains. How can we make it a subset? The "natural" way to do this is to set everything outside of <math>\mathcal S'</math> to zero. But that's exactly what <math>\mathcal V' = \{\mathbb V \in [0,1]^{\mathcal S} : \mathbb V(\phi) = 0 \text{ whenever }\phi \notin \mathcal S'\}</math> is. One thing I'm still not sure about is the "finite-dimensional" part; doesn't having <math>[0,1]</math> make the cube infinite-dimensional? | "Observe that <math>\mathcal V'</math> is equal to the natural inclusion of the finite-dimensional cube <math>[0,1]^{\mathcal S'}</math> in the space of all valuations <math>\mathcal V = [0,1]^{\mathcal S}</math>." -- I think what this is saying is that since <math>\mathcal S' \subset \mathcal S</math>, we can think of <math>[0,1]^{\mathcal S'}</math> as being sort of a subset of <math>[0,1]^{\mathcal S}</math>. Except it's not strictly speaking a subset, since the functions in <math>[0,1]^{\mathcal S'}</math> and <math>[0,1]^{\mathcal S}</math> have different domains. How can we make it a subset? The "natural" way to do this is to set everything outside of <math>\mathcal S'</math> to zero. But that's exactly what <math>\mathcal V' = \{\mathbb V \in [0,1]^{\mathcal S} : \mathbb V(\phi) = 0 \text{ whenever }\phi \notin \mathcal S'\}</math> is. One thing I'm still not sure about is the "finite-dimensional" part; doesn't having <math>[0,1]</math> make the cube infinite-dimensional? | ||
Definition of fix: I found it helpful to look at the [https://www.wolframalpha.com/input/?i=plot+f(x)+%3D+max(0,+min(1,+x)) graph] of <math>f(x) = \max(0, \min(1, x))</math>; this looks like the identity function in the interval <math>[0,1]</math>, but then becomes constant once it hits either of the endpoints. | |||
"the compact, convex space <math>\mathcal V'</math>" -- this intuitively makes sense, since <math>\mathcal V'</math> basically "looks like" a cube. But I'm not sure how to verify this. | "the compact, convex space <math>\mathcal V'</math>" -- this intuitively makes sense, since <math>\mathcal V'</math> basically "looks like" a cube. But I'm not sure how to verify this. |
Revision as of 18:32, 25 June 2019
Notes from the Logical Induction paper as I walk through the construction of LIA in section 5.
Lemma 5.1.1 (Fixed Point Lemma)
"Observe that is equal to the natural inclusion of the finite-dimensional cube in the space of all valuations ." -- I think what this is saying is that since , we can think of as being sort of a subset of . Except it's not strictly speaking a subset, since the functions in and have different domains. How can we make it a subset? The "natural" way to do this is to set everything outside of to zero. But that's exactly what is. One thing I'm still not sure about is the "finite-dimensional" part; doesn't having make the cube infinite-dimensional?
Definition of fix: I found it helpful to look at the graph of ; this looks like the identity function in the interval , but then becomes constant once it hits either of the endpoints.
"the compact, convex space " -- this intuitively makes sense, since basically "looks like" a cube. But I'm not sure how to verify this.
The following is used in the Fixed Point Lemma (5.1.1):
Writing the -strategy as
we have
But so the two sums cancel to obtain .