Interior and closure: what can observations settle?

Our four-card table uses X for dot and Y for stripe, in that bit order. A is one-zero, B one-one, C zero-one and D zero-zero. Classical X implies Y fails only when X is true and Y false. Its truth region collects the satisfying worlds. The positive observation opens describe what the chosen interface can certify. One means true and zero means false. Under positive observations X and Y, implication is true at B, C and D. Being inside this truth region answers a semantic question. Whether available positive certificates can establish membership is a different question, governed by the topology. Recall the four smallest neighborhoods: A has A and B; B has B alone; C has B and C; D has the whole carrier. Keep these sets beside E, the implication region B, C and D. We will test containment and intersection, rather than infer them from the drawing's geometry. A point is interior when some open neighborhood lies entirely inside E. In our finite space it is enough to test the smallest neighborhood. If it fits, it is a sufficient positive certificate; if it does not, no larger neighborhood will fit. B's singleton neighborhood fits inside the implication region. C's neighborhood, B and C, also fits. A's neighborhood contains A outside E. D's whole-carrier neighborhood also contains A, so those two points fail the interior test. The interior is B and C. In particular D satisfies implication but is not interior. That is a concrete instance of a true fact that this positive interface cannot certify at that point. A point belongs to the closure when every open neighborhood meets E. Since the smallest neighborhood is contained in all others, it suffices here to ask whether that smallest neighborhood meets E at all. Even A's neighborhood meets E through B. The other three points already lie in E, and each neighborhood contains its centre. Every point therefore passes the closure test, giving the whole carrier as the closure. If any open neighborhood witnesses interior membership, the smallest neighborhood is contained in it and also fits inside the event. Conversely the smallest neighborhood itself is open. That proves the interior criterion, rather than merely verifying our four examples. If the smallest neighborhood meets E, every containing neighborhood meets E. If it misses E, it is an explicit neighborhood proving failure of closure membership. These arguments depend on having smallest open neighborhoods, as our finite setting guarantees. Closure does not assign a likelihood, confirm that E is true, or model a delayed report. A can lie in the closure while falsifying implication. The claim is about the available neighborhoods and the inability of those neighborhoods to avoid E. The interior is also the union of all opens contained in E. This description works beyond our finite example. The neighborhood formula makes it especially easy to compute here, provided we preserve the correct information-order direction. Under positive X and Y, compute the interior and closure of the singleton containing A. Which smallest neighborhoods fit entirely inside it, and which merely meet it? No smallest neighborhood fits inside A alone, because even A's neighborhood also contains B. The interior is empty. A's and D's neighborhoods meet the singleton, while B's and C's do not, so its closure is A and D. Interior asks whether a whole neighborhood fits inside the event. Closure asks whether the neighborhood has any contact with it. The two tests have different quantifiers and can give very different answers on the same fixed truth region.