WEBVTT

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Our cards use X for dot and Y for stripe.

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

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D neither.

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A selected signature lists the truth answers in its formula slots.

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Its support lists the slot indices answered true.

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An available positive certificate affirms its formula; lack of a certificate does

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not certify negation.

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Begin with all four states and only the observation X.

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Its truth region contains A and B.

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The generated topology contains the empty set, that dot region, and the

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whole carrier.

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A and B have the same X signature.

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So do C and D.

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No open event from this topology separates A from B, or C

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from D.

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There is an open containing A but not D: the dot region.

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There is no open containing D but not A, because the only

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open containing D is the whole carrier.

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This is one-way separation.

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Describing the other two states as no-dot states does not make

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their region an available affirmative certificate.

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Now make not X available as a second observation.

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That gives the no-dot region its own affirmative certificate.

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All four states remain in the carrier; no premise has been learned.

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We now have four opens: the empty set, the dot block, the

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no-dot block and their union, the whole carrier.

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Intersecting the two blocks gives the empty set, so it adds nothing

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further.

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Each block is open and also closed: its complement is the other

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open block.

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A set that is both open and closed is called clopen.

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A topology is called discrete when every subset of its carrier is

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open.

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Does this make the four-state space discrete?

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In particular, can an open event contain A while excluding B?

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No.

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Every listed open contains both A and B or neither.

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The singleton containing A is not open.

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We separated the two blocks in both directions, but not the points

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inside either block.

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Start a new example on the same four-state carrier.

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Take the original list X, Y, and now make both signs available:

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X, not X, Y, and not Y.

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Consider the original signature pattern one, zero.

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Requiring X and not Y selects A alone.

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This conjunction is open in the enlarged menu and selects exactly that

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original signature class.

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This illustrates the general rule.

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Start with a finite list of formulas, then include a negation for

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every selected formula.

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A complete signature pattern can now be certified by requiring each true

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formula and the negation of each false one.

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That conjunction selects exactly the states with the original signature.

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Each original equal-signature class is therefore open.

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Unions of those classes are open too.

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Conversely, the new tests still agree on states that had identical original

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signatures.

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Intersections and unions cannot split such a class.

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So the opens are exactly the unions of the original classes.

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This is a partition topology.

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It is discrete on the original states precisely when every original class

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contains just one state.

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Adding both signs does not recover a distinction the original list never

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recorded.

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Return now to the original positive list X, Y on the same

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four-state carrier.

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We can compare separation levels in this space.

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T zero asks whether every distinct pair has an open containing one

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point and excluding the other.

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At least one direction is enough.

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Equivalently here, different states must have different signatures.

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A map is injective when different inputs give different outputs.

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So the observation space is T zero exactly when its signature map

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is injective.

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Under X, Y, all four signatures differ.

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The space is T zero.

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But look at A and B.

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The Y region contains B without A.

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No open contains A without B.

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T one requires both directions for every distinct pair.

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It need not be the same pair of open sets for all

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comparisons.

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Our positive X, Y space fails T one at A and B,

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even though it is T zero.

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A and C can be separated both ways: X contains A without

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C, and Y contains C without A.

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One successful pair does not establish T one for the whole space.

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Here is why the finite assumption matters.

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Fix one point v in a finite T one space.

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For every other point, choose an open containing v but excluding that

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other point.

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Intersect all those opens.

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The intersection still contains v, and every other point has been excluded.

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So the intersection is the singleton containing v.

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There are only finitely many other points, and a topology keeps finite

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intersections open.

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Every singleton is therefore open.

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Any subset is a union of singletons, so every subset is open:

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the topology is discrete.

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If the carrier has just one point, its singleton is already the

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whole open carrier.

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An empty carrier is discrete too.

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The finite-intersection argument does not justify the same conclusion for an

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arbitrary infinite space.

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The signature criterion for T one has two parts.

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First, the signature map must be injective.

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Second, its realized supports must form an antichain: no two different supports

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are contained in one another.

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Why does containment matter?

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If every selected formula true at A is also true at B,

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any finite positive conjunction true at A remains true at B.

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The same is true of unions of such conjunctions.

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No generated open can contain A without B.

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Conversely, when A's support is not contained in B's, some

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selected formula is true at A and false at B.

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Its truth region is an open witnessing that direction.

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Requiring this both ways gives exactly the criterion.

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Under X, Y the map is injective, but A's support is

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contained in B's, so the antichain condition fails.

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Under X, not X the two realized supports form an antichain, but

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A and B share a support, so injectivity fails.

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Both conditions are necessary.

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Now change the carrier explicitly.

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Use just one variable X, so the states are zero and one.

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This is a separate two-point example, not a premise update of

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the four-card universe.

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With only X, the opens are empty, the singleton containing one, and

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the whole two-point carrier.

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This is the Sierpinski example: T zero but not T one.

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Add not X, and the singleton containing zero also becomes open.

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All four subsets are now open, so this two-point space is

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discrete.

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On our original four-state carrier, the same test names leave A

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and B together, and C and D together.

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The carrier label is what makes the two conclusions consistent.

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Return to the four-state carrier.

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Compare X alone; X with not X; positive X and Y; and

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both signs of X and Y.

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Which gives T zero?

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Which gives T one?

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Which is discrete?

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X alone is neither T zero nor T one, and is not

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discrete.

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Adding not X gives a partition topology, but A and B still

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share a signature, so it is still neither T zero nor T

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one.

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Positive X and Y is T zero, but not T one or

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discrete.

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Finally, both signs of X and Y make each complete description available.

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X and not Y selects A; X and Y selects B; not

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X and Y selects C; not X and not Y selects D.

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Every singleton is now open.

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The topology is discrete and therefore T one and T zero.

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Two-sided observations open the original signature classes; full separation needs those

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classes to be singletons.

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Next we will arrange positive-information comparisons into an order, fixing the

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direction carefully.

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Distinguishability, two-way separation and discreteness will remain different questions.