False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)

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FALSE: every compact space is sequentially compact

Statement

False claim: every compact topological space (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right) is sequentially compact (Countably compact, Lindel"of, sequentially compact, limit point compact and $\sigma$ -compact spaces, and relatively compact subsets).

Where the claim comes from, and what is actually true. For a metric space the two conditions are equivalent, and the claim above is that equivalence transplanted to an arbitrary topological space. The refutation builds its own witness out of Tychonoff's theorem (Tychonoff's theorem: an arbitrary product of compact spaces is compact in the product topology, assuming the Axiom of Choice): the product

$Y \;:=\; \prod_{x \in D} \{0,1\}, \qquad D := \{0,1\}^{\mathbb{N}},$

of one copy of the two-point discrete space for every $0$ - $1$ sequence, together with the sequence $(F_n)$ in $Y$ whose $n$ -th term reads off the $n$ -th coordinate, $F_n(x) := x_n$ . The Axiom of Choice is assumed, since Tychonoff's theorem: an arbitrary product of compact spaces is compact in the product topology, assuming the Axiom of Choice carries it.

Facts & Assumptions

Given: The two-point discrete space $\{0,1\}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), the set $D = \{0,1\}^{\mathbb{N}}$ of functions $\mathbb{N} \to \{0,1\}$ (The natural numbers $\mathbb{N}$ (von Neumann)), the product $Y = \prod_{x \in D} \{0,1\}$ with the product topology (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space), the projections $\pi_x : Y \to \{0,1\}$ , and the elements $F_n \in Y$ defined by $F_n(x) := x_n$ for $n \in \mathbb{N}$ and $x \in D$ .

[A1]

The false claim: every compact topological space is sequentially compact.

[L1]

Every finite space is compact, so $\{0,1\}$ with the discrete topology is compact; and a product of compact spaces is compact in the product topology, assuming the Axiom of Choice (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Tychonoff's theorem: an arbitrary product of compact spaces is compact in the product topology, assuming the Axiom of Choice).

[L2]

The sets $\pi_x^{-1}[U]$ with $x \in D$ and $U \subseteq \{0,1\}$ are open in $Y$ , being members of the subbasis of the product topology, every subset of the discrete $\{0,1\}$ being open (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space, The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L3]

A sequence in a space is a function on $\mathbb{N}$ ; it converges to $p$ when every open set containing $p$ contains all but finitely many of its terms; a subsequence is given by a strictly increasing index map $j \mapsto n_j$ , which satisfies $n_j \ge j$ ; and a space is sequentially compact when every sequence has a convergent subsequence (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure, Sequences of reals: bounded, eventually, frequently, tails, subsequences, A strictly increasing index map satisfies $n_k \ge k$ , Countably compact, Lindel"of, sequentially compact, limit point compact and $\sigma$ -compact spaces, and relatively compact subsets).

Refutation

technique · contradiction

1.1

Suppose the claim [A1] holds, so that every compact space is sequentially compact.

A1assume-contra

1.2

$Y$ is compact by [L1], being a product of copies of the compact two-point discrete space.

2.1

By [A1] and step 1.2 the sequence $(F_n)$ in $Y$ has a subsequence $(F_{n_j})$ converging to some $G \in Y$ , the index map $j \mapsto n_j$ being strictly increasing.

A1L3step 1.2

3.1

Define $x \in D$ by $x_k := 0$ when $k = n_j$ for an even $j$ , $x_k := 1$ when $k = n_j$ for an odd $j$ , and $x_k := 0$ for every $k$ not of the form $n_j$ ; this is well defined because $j \mapsto n_j$ is injective, being strictly increasing. Then $F_{n_j}(x) = x_{n_j}$ is $0$ for even $j$ and $1$ for odd $j$ .

L3step 2.1construct

4.1

The set $\pi_x^{-1}[\{G(x)\}]$ is open by [L2] and contains $G$ , so by step 2.1 it contains $F_{n_j}$ for all large $j$ ; that is, $F_{n_j}(x) = G(x)$ for all large $j$ . But step 3.1 makes $F_{n_j}(x)$ take the value $0$ at every even $j$ and $1$ at every odd $j$ , so it is constant on no set of large indices. This contradiction refutes the claim [A1].

A1L2step 2.1step 3.1discharge-contradiction∎

Remarks

What the witness exploits. Compactness of a product is a statement about covers and survives an index set of any size; sequential compactness is a statement about countably many terms and does not. The index set $D$ here is the set of all $0$ - $1$ sequences, and the point $x$ built at step 3.1 is chosen to disagree with the given subsequence at exactly the places that matter, which is possible precisely because every $0$ - $1$ sequence is available as an index.

No binary expansion of a real number is used, and none is needed: the witness is built from $\{0,1\}$ -valued functions directly, so nothing here rests on the representation of reals by digits.

The Axiom of Choice is assumed only through Tychonoff's theorem: an arbitrary product of compact spaces is compact in the product topology, assuming the Axiom of Choice, which is where compactness of $Y$ comes from. Nothing else in the refutation selects anything; the point $x$ is defined by a rule.

Depends on

Used by

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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 123 results over 29 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Sequentially compact space (Wikipedia) (standard reference, not scraped)
Tychonoff's theorem (Wikipedia) (standard reference, not scraped)
Stacks Project, Tag 08ZU (standard reference, not scraped)