WEBVTT

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An inverse image consists of the source points whose outputs land in

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a chosen target event.

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Continuity requires every open target event to have an open inverse image.

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Our specialization direction says that every open containing v also contains w;

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for observation spaces, w retains every selected positive true at v.

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Continuity preserves the chosen specialization direction.

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Suppose every source open containing v contains w.

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To compare their images, start with any target open containing the image

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of v and pull it backward.

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Continuity makes that inverse image source-open.

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Since it contains v, it contains w.

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Therefore the original target open contains the image of w.

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Every target neighborhood of the first image contains the second, proving the

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comparison.

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A map that preserves these comparisons is called monotone.

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In our finite spaces, every open is upward closed.

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If a source point lies in the inverse image of a target

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open, every source point above it maps above its image.

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Monotonicity keeps those higher images inside the target open.

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Its inverse image is therefore upward closed and hence open in the

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finite source topology.

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This proves continuity.

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The finite upper-set description is the assumption making this converse available

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here.

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Give source and target the positive X,Y topology.

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For this map, each side uses the original four X,Y assignments:

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A is one-zero, B one-one, C zero-one and D

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zero-zero.

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The tests are positive X and Y.

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Read both source and target labels using that same two-bit key.

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One means true and zero means false.

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Send a valuation to two copies of the disjunction of its bits.

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D maps to D, while A, B and C map to B.

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Display the map explicitly before attaching any continuity verdict.

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D maps to the bottom point and all the nonzero points map

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to the top point.

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Every ordered pair remains ordered after applying the map.

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The map is monotone and therefore continuous, although three distinct points share

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an image.

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Under unit coordinate weights, D and A differ only on X, so

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their symmetric distance is one.

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These are the original source points, not their images.

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Keep the two comparisons on separate labelled rows.

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Their images are D and B, which differ on both coordinates.

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The target distance is two, compared with source distance one.

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A map is nonexpansive when no pair's distance increases.

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This continuous map fails that condition, and it also fails injectivity because

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several inputs share an output.

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One can package all selected bit maps into a tuple of Sierpinski-

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valued coordinates.

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The product topology is generated by conditions on individual coordinates.

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Pulling those conditions back gives exactly the selected truth-region generators.

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The observation topology is the smallest source topology making every selected bit

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coordinate continuous.

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This is the initial-topology description.

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It recovers the same space; it does not recover an arbitrary list

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of weights or a unique observation presentation.

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Does continuity guarantee that no pair's symmetric distance increases?

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Use the displayed map and the pair D, A.

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State separately what continuity does preserve and what this example refutes.

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The map preserves every positive-information comparison, but it changes this distance

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from one to two.

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Continuity preserves the qualitative order in these finite spaces; the quantitative distances

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contain additional presentation and weight data.

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We now have two compatible tests: pull back target opens, or preserve

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specialization comparisons in these finite upper-set spaces.

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Neither test asserts preservation of identities, distances or all distinctions between source

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points.