One-sided and two-sided observation

Our cards use X for dot and Y for stripe. A has only a dot, B both, C only a stripe, and D neither. A selected signature lists the truth answers in its formula slots. Its support lists the slot indices answered true. An available positive certificate affirms its formula; lack of a certificate does not certify negation. Begin with all four states and only the observation X. Its truth region contains A and B. The generated topology contains the empty set, that dot region, and the whole carrier. A and B have the same X signature. So do C and D. No open event from this topology separates A from B, or C from D. There is an open containing A but not D: the dot region. There is no open containing D but not A, because the only open containing D is the whole carrier. This is one-way separation. Describing the other two states as no-dot states does not make their region an available affirmative certificate. Now make not X available as a second observation. That gives the no-dot region its own affirmative certificate. All four states remain in the carrier; no premise has been learned. We now have four opens: the empty set, the dot block, the no-dot block and their union, the whole carrier. Intersecting the two blocks gives the empty set, so it adds nothing further. Each block is open and also closed: its complement is the other open block. A set that is both open and closed is called clopen. A topology is called discrete when every subset of its carrier is open. Does this make the four-state space discrete? In particular, can an open event contain A while excluding B? No. Every listed open contains both A and B or neither. The singleton containing A is not open. We separated the two blocks in both directions, but not the points inside either block. Start a new example on the same four-state carrier. Take the original list X, Y, and now make both signs available: X, not X, Y, and not Y. Consider the original signature pattern one, zero. Requiring X and not Y selects A alone. This conjunction is open in the enlarged menu and selects exactly that original signature class. This illustrates the general rule. Start with a finite list of formulas, then include a negation for every selected formula. A complete signature pattern can now be certified by requiring each true formula and the negation of each false one. That conjunction selects exactly the states with the original signature. Each original equal-signature class is therefore open. Unions of those classes are open too. Conversely, the new tests still agree on states that had identical original signatures. Intersections and unions cannot split such a class. So the opens are exactly the unions of the original classes. This is a partition topology. It is discrete on the original states precisely when every original class contains just one state. Adding both signs does not recover a distinction the original list never recorded. Return now to the original positive list X, Y on the same four-state carrier. We can compare separation levels in this space. T zero asks whether every distinct pair has an open containing one point and excluding the other. At least one direction is enough. Equivalently here, different states must have different signatures. A map is injective when different inputs give different outputs. So the observation space is T zero exactly when its signature map is injective. Under X, Y, all four signatures differ. The space is T zero. But look at A and B. The Y region contains B without A. No open contains A without B. T one requires both directions for every distinct pair. It need not be the same pair of open sets for all comparisons. Our positive X, Y space fails T one at A and B, even though it is T zero. A and C can be separated both ways: X contains A without C, and Y contains C without A. One successful pair does not establish T one for the whole space. Here is why the finite assumption matters. Fix one point v in a finite T one space. For every other point, choose an open containing v but excluding that other point. Intersect all those opens. The intersection still contains v, and every other point has been excluded. So the intersection is the singleton containing v. There are only finitely many other points, and a topology keeps finite intersections open. Every singleton is therefore open. Any subset is a union of singletons, so every subset is open: the topology is discrete. If the carrier has just one point, its singleton is already the whole open carrier. An empty carrier is discrete too. The finite-intersection argument does not justify the same conclusion for an arbitrary infinite space. The signature criterion for T one has two parts. First, the signature map must be injective. Second, its realized supports must form an antichain: no two different supports are contained in one another. Why does containment matter? If every selected formula true at A is also true at B, any finite positive conjunction true at A remains true at B. The same is true of unions of such conjunctions. No generated open can contain A without B. Conversely, when A's support is not contained in B's, some selected formula is true at A and false at B. Its truth region is an open witnessing that direction. Requiring this both ways gives exactly the criterion. Under X, Y the map is injective, but A's support is contained in B's, so the antichain condition fails. Under X, not X the two realized supports form an antichain, but A and B share a support, so injectivity fails. Both conditions are necessary. Now change the carrier explicitly. Use just one variable X, so the states are zero and one. This is a separate two-point example, not a premise update of the four-card universe. With only X, the opens are empty, the singleton containing one, and the whole two-point carrier. This is the Sierpinski example: T zero but not T one. Add not X, and the singleton containing zero also becomes open. All four subsets are now open, so this two-point space is discrete. On our original four-state carrier, the same test names leave A and B together, and C and D together. The carrier label is what makes the two conclusions consistent. Return to the four-state carrier. Compare X alone; X with not X; positive X and Y; and both signs of X and Y. Which gives T zero? Which gives T one? Which is discrete? X alone is neither T zero nor T one, and is not discrete. Adding not X gives a partition topology, but A and B still share a signature, so it is still neither T zero nor T one. Positive X and Y is T zero, but not T one or discrete. Finally, both signs of X and Y make each complete description available. X and not Y selects A; X and Y selects B; not X and Y selects C; not X and not Y selects D. Every singleton is now open. The topology is discrete and therefore T one and T zero. Two-sided observations open the original signature classes; full separation needs those classes to be singletons. Next we will arrange positive-information comparisons into an order, fixing the direction carefully. Distinguishability, two-way separation and discreteness will remain different questions.