ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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The coordinate-reading sequence in a compact binary cube has a convergent subnet but no convergent subsequence

Example

Let $D = {0, 1}^{N}$ and $Y = {0, 1}^{D}$ with the product topology. The coordinate-reading sequence is $F_{n} (r) = r_{n}$ . Assuming the ultrafilter lemma, $Y$ is compact and $(F_{n})$ has a convergent subnet, but it has no convergent subsequence.

Facts & Assumptions

Given: The binary cube and the coordinate-reading sequence above.

[L1]

The published refutation FALSE: every compact space is sequentially compact defines this cube and sequence as a compact nonsequentially compact witness.

[L2]

Under the ultrafilter lemma, every net in a compact space has a convergent subnet (Assuming the ultrafilter lemma, compactness is equivalent to every net having a cluster point, every net having a convergent subnet, every filter having a cluster point, and every ultrafilter converging).

[L3]

Under the ultrafilter lemma, an arbitrary product of compact Hausdorff spaces is compact (Assuming the ultrafilter lemma, an arbitrary product of compact Hausdorff spaces is compact).

[L4]

Each coordinate projection from a product is continuous, so it sends a convergent net to a convergent coordinate net (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, A map of topological spaces is continuous at a point if and only if it preserves every net converging to that point).

Verification

technique · contradiction

1.1

Each two-point discrete factor is compact and Hausdorff, so the cube is compact under the ultrafilter lemma by [L3]; then [L2] gives a convergent subnet of $(F_{n})$ .

L2L3

1.2

Assume for a contradiction that $F_{n_{j}}$ is a convergent subsequence. Define $r \in D$ by $r_{n_{j}} = 0$ for even $j$ and $r_{n_{j}} = 1$ for odd $j$ , assigning $0$ elsewhere.

L1assume-contra

2.1

The $r$ -coordinate of $F_{n_{j}}$ alternates $0, 1$ , so it does not converge in the discrete two-point factor. By [L4], a convergent product net has convergent coordinate nets, contradiction.

step 1.2L4

3.1

Hence no convergent subsequence exists, while step 1.1 supplies a convergent subnet.

step 1.1step 2.1discharge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

53 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Product topology (Wikipedia) (standard reference, not scraped)
Compact space (Wikipedia) (standard reference, not scraped)
Tychonoff's theorem (Wikipedia) (standard reference, not scraped)
Sequential space (Wikipedia) (standard reference, not scraped)