WEBVTT

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On the original cards, X means dot and Y means stripe: A

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has only a dot, B both, C only a stripe, and D

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neither.

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A selected signature records the truth answers to the chosen formula slots.

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Open events are generated from their true-row regions using unions and

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finite intersections.

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Selecting a menu has not deleted any card.

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Let only X be observable.

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The generated opens are the empty set, the dot region containing A

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and B, and the whole carrier.

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A and B are different valuations, yet they belong to exactly the

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same open sets.

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C and D also belong to exactly the same opens.

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Topological indistinguishability means this equality of open-set membership.

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It does not assert that the points are the same valuation or

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that all formulas have the same truth value there.

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Equal signatures mean agreement on each selected formula, hence on each subbasic

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truth region.

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If two points agree on each generator, they agree on every finite

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intersection of generators.

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They then agree on every union of those intersections.

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Those unions are all the generated opens.

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Equal signatures therefore imply topological indistinguishability.

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This argument explains why checking the selected coordinates is sufficient; we have

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not merely checked a few opens and guessed the rest.

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Conversely, if two points have the same membership in every open, they

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agree on each subbasic open in particular.

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Each selected truth value agrees, so their signatures are equal.

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The argument runs in both directions.

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For the menu X we can replace the indistinguishable pair A and

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B by a class labelled P, and C and D by a

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class labelled Q.

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These are two new class labels, not surviving individual valuations.

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The dot region becomes the singleton class P, and the whole carrier

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becomes both classes.

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Together with the empty set, these are the quotient opens.

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A set of classes is open exactly when all the original points

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in those classes form an open set before the quotient.

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The quotient distinguishes its two points in one direction, but the singleton

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Q is still not open.

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Merging indistinguishable points does not force all subsets open.

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This two-point quotient has the same three-open shape as the

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Sierpinski example.

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A space is T zero if any two distinct points differ in

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membership in at least one open.

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One direction of separation is enough.

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For an observation space, this is equivalent to different valuations having different

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signatures.

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A map with different outputs for different inputs is injective.

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Thus the original observation space is T zero exactly when its signature

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map is injective.

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On the quotient, different classes already have different realized signatures, so the

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quotient is T zero.

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With only X and Y as one conjunction observation, which points merge?

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What are the quotient opens?

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Does the quotient automatically become discrete?

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B has the true one-entry signature.

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A, C and D have the false one-entry signature.

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Those are the two classes.

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The quotient opens are empty, the class containing B, and both classes;

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the false class alone is absent.

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The quotient removes duplicate descriptions.

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It does not remove every one-way information comparison.

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Later, a symmetric metric on the same class set will give a

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different, discrete topology; the underlying quotient set alone does not specify a

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topology.