TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22

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The compact T1 product theorem is equivalent to AC

Statement

Over $Z F$ , the Axiom of Choice (The Axiom of Choice) is equivalent to the assertion that every product of compact $T_{1}$ spaces ( $T_{0}$ (Kolmogorov) and $T_{1}$ (Frechet) spaces, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, 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) is compact.

Facts & Assumptions

Given: A family $(A_{i})_{i \in I}$ of nonempty sets; the repaired coordinates $X_{A_{i}}$ of The isolated-point repair of Kelley's choice space; the product $X : = \prod_{i} X_{A_{i}}$ with projections $π_{i}$ .

[F1]

Each $X_{A_{i}}$ is compact and $T_{1}$ , and $A_{i}$ is a closed subspace of it (The isolated-point repair of Kelley's choice space, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

[F2]

Under AC, Tychonoff's theorem gives compactness of arbitrary products of compact spaces (Tychonoff's theorem: an arbitrary product of compact spaces is compact in the product topology, assuming the Axiom of Choice, The Axiom of Choice).

[F3]

A space is compact if and only if every family of closed sets with the finite intersection property has nonempty intersection (A space is compact exactly when every family of closed subsets with the finite intersection property has nonempty intersection, Finite intersection property).

[F4]

If $n \in N$ and $F : n \to V$ is a natural-number-indexed list of nonempty sets, then the family of values $F [n]$ has a choice function (Every natural-number-indexed list of nonempty sets has a choice function on its family of values).

[L1]

A cylinder $π_{i}^{- 1} [A_{i}]$ is closed in $X$ when $A_{i}$ is closed in $X_{A_{i}}$ , being the preimage of a closed set under a continuous projection (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, For $A \subseteq S \subseteq X$ the closure of $A$ in $S$ is ${\overline{A}}^{X} \cap S$ , while the interior only contains ${int}^{X} (A) \cap S$ , with equality when $S$ is open; and a dense subset of $X$ traces to a dense subset of every open $S$ ).

Proof

technique · direct

1.1assume-hypF2

Under AC every product of compact spaces is compact by [F2], so every product of compact $T_{1}$ spaces is compact; this is the forward direction.

1.2assume-hypgiven

Conversely, assume every product of compact $T_{1}$ spaces is compact, and let $(A_{i})_{i \in I}$ be a family of nonempty sets; if $I = \emptyset$ the product over the empty index set is a one-point space and the unique element is a choice function, and if $I \neq \emptyset$ we build one below.

2.1step 1.2F1

Form $X : = \prod_{i \in I} X_{A_{i}}$ where $X_{A_{i}}$ is the repaired coordinate of [F1]; each factor is compact $T_{1}$ , so $X$ is compact by the hypothesis.

3.1step 2.1F1F4L1

For each $i$ the cylinder $C_{i} : = π_{i}^{- 1} [A_{i}]$ is closed in $X$ by [L1] and [F1]. To verify the finite intersection property in its finite-list form, let $n \in N$ and let $s : n \to {C_{i} : i \in I}$ be arbitrary. For each $k < n$ , choose the unique $i_{k} \in I$ with $s (k) = C_{i_{k}}$ (the cylinder determines its coordinate), and define $F (k) : = A_{i_{k}}$ . By [F4] the family $F [n]$ has a choice function $h$ . Define $x \in X$ by $x (i) : = h (A_{i})$ when $i = i_{k}$ for some $k < n$ , and by the distinguished point $\infty \in X_{A_{i}}$ otherwise. If the same coordinate occurs more than once this gives the same value, and $h (A_{i}) \in A_{i}$ ; hence $x \in C_{i_{k}} = s (k)$ for every $k < n$ . Thus every finite list from ${C_{i} : i \in I}$ has nonempty intersection.

4.1step 2.1step 3.1F3

By compactness of $X$ and [F3] the intersection $⋂_{i} C_{i}$ is nonempty; any point $x$ of it has $x_{i} \in A_{i}$ for every $i$ , so $i \mapsto x_{i}$ is a choice function for the family $(A_{i})_{i \in I}$ .

5.1step 1.1step 4.1∎

The family of nonempty sets was arbitrary, so AC holds; together with step 1.1 this proves the displayed equivalence.

The compact T1 product theorem is equivalent to AC

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources