WEBVTT 00:00:00.250 --> 00:00:02.996 We are continuing the open-set interpretation. 00:00:03.625 --> 00:00:06.474 Interior means the largest open subset. 00:00:06.796 --> 00:00:10.822 Open-set negation takes the interior of the complement each time that 00:00:10.822 --> 00:00:12.480 connective is evaluated. 00:00:12.941 --> 00:00:16.250 Earlier classical negation used the complement itself. 00:00:16.860 --> 00:00:21.521 Keep that distinction while we introduce the corresponding open implication. 00:00:22.495 --> 00:00:27.172 Given open events U and V, seek an open condition W such 00:00:27.172 --> 00:00:29.972 that W together with U guarantees V. 00:00:30.651 --> 00:00:35.524 Among all such open conditions, choose the largest by inclusion. 00:00:36.047 --> 00:00:41.408 This is the least restrictive sufficient open condition: it allows as many 00:00:41.440 --> 00:00:44.621 points as possible while retaining the guarantee. 00:00:45.567 --> 00:00:50.602 Without the openness requirement, a point is allowed whenever it lies outside 00:00:50.612 --> 00:00:52.066 U or inside V. 00:00:52.720 --> 00:00:55.397 This is the classical implication region. 00:00:55.954 --> 00:01:00.978 But that region may not be open in the selected observation topology. 00:01:02.087 --> 00:01:07.942 Taking the interior supplies the largest open subset of that classical candidate. 00:01:08.341 --> 00:01:13.245 It is therefore the largest open W whose intersection with U lies 00:01:13.267 --> 00:01:13.733 in V. 00:01:14.537 --> 00:01:18.415 The operation is called Heyting implication. 00:01:18.687 --> 00:01:21.356 Here is the defining equivalence in words. 00:01:22.015 --> 00:01:27.256 For any open W, W is contained in the Heyting implication from 00:01:27.311 --> 00:01:32.356 U to V if and only if the intersection of W with 00:01:32.378 --> 00:01:34.092 U is contained in V. 00:01:34.791 --> 00:01:38.149 The openness of W is part of this rule. 00:01:38.559 --> 00:01:42.965 The two sides express the same guarantee in different positions. 00:01:44.114 --> 00:01:46.598 Use the positive four-state topology. 00:01:47.603 --> 00:01:52.411 On this four-card carrier, the fixed X,Y values are A: 00:01:52.649 --> 00:01:58.936 one-zero, B: one-one, C: zero-one and D: zero-zero. 00:01:59.458 --> 00:02:02.701 Only positive X and Y are selected. 00:02:03.150 --> 00:02:08.066 The smallest neighborhoods are A and B from A; B alone from 00:02:08.076 --> 00:02:11.972 B; B and C from C; and the whole carrier from D. 00:02:12.731 --> 00:02:15.819 One means true and zero means false. 00:02:16.664 --> 00:02:20.123 U is the X region and V is the Y region. 00:02:20.650 --> 00:02:25.391 The classical implication candidate is the complement of U together with V, 00:02:25.775 --> 00:02:27.620 the set B, C and D. 00:02:28.924 --> 00:02:31.962 Its largest open subset is B and C. 00:02:32.700 --> 00:02:37.094 D cannot be included in any open subset of the candidate because 00:02:37.094 --> 00:02:41.564 D's smallest neighborhood is the whole carrier, including A outside the 00:02:41.564 --> 00:02:42.300 candidate. 00:02:42.816 --> 00:02:46.754 The interior removes exactly that obstruction here. 00:02:47.486 --> 00:02:52.699 The implication itself is an open condition satisfying the defining guarantee. 00:02:53.142 --> 00:02:56.527 Intersect it with U and the result is included in V. 00:02:57.276 --> 00:03:02.629 This is open-set Modus Ponens, obtained directly from the largest-condition 00:03:02.650 --> 00:03:03.470 property. 00:03:04.330 --> 00:03:08.403 In our example the intersection is B, which lies in V. 00:03:09.102 --> 00:03:15.587 The visible calculation illustrates the general inclusion; the defining property, rather than 00:03:15.587 --> 00:03:20.438 this one four-point check, explains why the rule always holds for 00:03:20.459 --> 00:03:21.566 open inputs. 00:03:22.690 --> 00:03:26.370 Combine the implication with the open negation of V. 00:03:27.034 --> 00:03:31.560 If this open condition also met U, Modus Ponens would put the 00:03:31.571 --> 00:03:36.107 intersection in V, while the negation keeps it disjoint from V. 00:03:36.841 --> 00:03:39.511 The intersection with U must be empty. 00:03:40.625 --> 00:03:45.029 Being an open condition disjoint from U, it is included in the 00:03:45.061 --> 00:03:49.333 largest such open condition, Heyting negation of U. 00:03:49.888 --> 00:03:56.181 This derives the open-set Modus Tollens inequality without replacing Heyting negation 00:03:56.301 --> 00:03:58.531 by an arbitrary complement. 00:03:59.299 --> 00:04:04.009 Why can we not add D to the computed Heyting implication while 00:04:04.009 --> 00:04:06.892 keeping it open and retaining the guarantee? 00:04:07.456 --> 00:04:11.817 Use D's smallest neighborhood rather than the visual position of its 00:04:11.829 --> 00:04:12.317 card. 00:04:30.620 --> 00:04:36.264 Every open containing D contains the whole carrier in this positive topology. 00:04:36.775 --> 00:04:41.731 It would therefore include A, which lies in U but outside V. 00:04:42.480 --> 00:04:44.141 The guarantee fails. 00:04:44.765 --> 00:04:48.563 The largest allowed open is B and C, with no D. 00:04:49.937 --> 00:04:56.288 The opens with intersection, union and these implication and negation operations form 00:04:56.310 --> 00:04:58.178 a Heyting algebra. 00:04:58.450 --> 00:05:03.951 In a discrete topology the interiors do nothing and these operations recover 00:05:03.973 --> 00:05:05.591 the classical set forms. 00:05:06.147 --> 00:05:10.690 Keep the two semantics explicitly labelled in every comparison.