Indistinguishable states and the T0 quotient

On the original cards, X means dot and Y means stripe: A has only a dot, B both, C only a stripe, and D neither. A selected signature records the truth answers to the chosen formula slots. Open events are generated from their true-row regions using unions and finite intersections. Selecting a menu has not deleted any card. Let only X be observable. The generated opens are the empty set, the dot region containing A and B, and the whole carrier. A and B are different valuations, yet they belong to exactly the same open sets. C and D also belong to exactly the same opens. Topological indistinguishability means this equality of open-set membership. It does not assert that the points are the same valuation or that all formulas have the same truth value there. Equal signatures mean agreement on each selected formula, hence on each subbasic truth region. If two points agree on each generator, they agree on every finite intersection of generators. They then agree on every union of those intersections. Those unions are all the generated opens. Equal signatures therefore imply topological indistinguishability. This argument explains why checking the selected coordinates is sufficient; we have not merely checked a few opens and guessed the rest. Conversely, if two points have the same membership in every open, they agree on each subbasic open in particular. Each selected truth value agrees, so their signatures are equal. The argument runs in both directions. For the menu X we can replace the indistinguishable pair A and B by a class labelled P, and C and D by a class labelled Q. These are two new class labels, not surviving individual valuations. The dot region becomes the singleton class P, and the whole carrier becomes both classes. Together with the empty set, these are the quotient opens. A set of classes is open exactly when all the original points in those classes form an open set before the quotient. The quotient distinguishes its two points in one direction, but the singleton Q is still not open. Merging indistinguishable points does not force all subsets open. This two-point quotient has the same three-open shape as the Sierpinski example. A space is T zero if any two distinct points differ in membership in at least one open. One direction of separation is enough. For an observation space, this is equivalent to different valuations having different signatures. A map with different outputs for different inputs is injective. Thus the original observation space is T zero exactly when its signature map is injective. On the quotient, different classes already have different realized signatures, so the quotient is T zero. With only X and Y as one conjunction observation, which points merge? What are the quotient opens? Does the quotient automatically become discrete? B has the true one-entry signature. A, C and D have the false one-entry signature. Those are the two classes. The quotient opens are empty, the class containing B, and both classes; the false class alone is absent. The quotient removes duplicate descriptions. It does not remove every one-way information comparison. Later, a symmetric metric on the same class set will give a different, discrete topology; the underlying quotient set alone does not specify a topology.