Tychonoff's Theorem Without Choice

== Theorem without Axiom of Choice ==
=== Preliminaries ===
Let $\ds X = \prod_{i \mathop \in I} X_i$ be a topological product space.
From the definition of the product topology, a basic open set of the natural basis of $X$ is a set of the form:
:$\ds \prod_{i \mathop \in I} U_i$
where:
:each $U_i$ is a nonempty open subset of $X_i$
and:
:$U_i = X_i$ for all but finitely many $i \in I$.
=== Statement that holds in Zermelo–Fraenkel set theory (without Axiom of Choice) ===
Let $\struct {I, <}$ be a well-ordered set.
Zermelo's Well-Ordering Theorem, equivalent to the Axiom of Choice over Zermelo–Fraenkel set theory, states that every set is well-orderable.
Let $\family {X_i}_{i \mathop \in I}$ be an indexed family of compact topological spaces.
Denote $\ds X := \prod_{i \mathop \in I} X_i$.
Let $\struct {F, \subset}$ be the set-theoretic tree:
:$\ds \paren {\bigcup_{i \mathop \in I} \prod_{j \mathop < i} X_j} \cup X$
of mappings defined on initial intervals of $I$, where the ordering is that of set inclusion.
Consider the set of all subtrees $T \subseteq F$ with the following property:
:For every $i \in I$ and every $\ds f \in T \cap \prod_{j \mathop < i} X_j$, the set $\set {\map g i: g \in T, f \subsetneq g}$ is closed in $X_i$.
Suppose that every such subtree $T$ of $F$ has a branch.
Hausdorff's Maximal Principle, equivalent to the Axiom of Choice over Zermelo–Fraenkel set theory, states that in an ordered set, every chain is contained in a maximal chain, which implies that every tree has a branch.
Then $\ds \prod_{i \mathop \in I} X_i$ is compact.
== Proof of the version without Axiom of Choice ==
To prove that every open cover of $X$ has a finite subcover, it is enough to prove that every open cover by basic open set of the natural basis has a finite subcover.
(If the Axiom of Choice was assumed, it would be even enough to prove that every open cover by sub-basic open sets of the natural sub-basis has a finite subcover, according to Alexander's Sub-Basis Theorem.)
Let $\OO$ be a collection of basic open subsets of $X$ such that no finite subcollection of $\OO$ covers $X$.
It is enough to prove that $\OO$ does not cover $X$.
Let $T_\OO$ be the set of all elements $f \in F$ such that the set $\set {x \in X : f \subseteq x}$ is not covered by any finite subcollection of $\OO$.
Then $T_\OO$ is a subtree of $F$.
For every $i \in I$ and $\ds f \in T_\OO \cap \prod_{j \mathop < i} X_j$, denote:
:$\map {C_\OO} f = \set {\map g i: f \subsetneq g \in T_\OO}$
=== Step 1 ===
For every $i \in I$ and every $\ds f \in T_\OO \cap \prod_{j \mathop < i} X_j$:
:$\map {C_\OO} f$ is closed in $X_i$ and nonempty.
Here the compactness of $X_i$ is used similarly to the usual proof: Topological Product of Compact Spaces.
To prove this, let $\WW$ be the set of all open subsets $U$ of $X_i$ such that:
:there exist a finite subset $\PP \subseteq \OO$ such that:
::$\ds \set {x \in X: f \subseteq x, \map x i \in U} \subseteq \bigcup \PP$
Then:
:$\ds X_i \setminus \map {C_\OO} f = \bigcup \WW$
and hence $\map {C_\OO} f$ is closed.
 $\map {C_\OO} f$ is empty.
Then $\WW$ would be a cover for $X_i$ and, by compactness of $X_i$, it would have a finite subcover.
This would yield a finite subset of $\OO$ that covers $\set {x \in X: f \subseteq x}$ in contradiction with the fact that $f \in T_\OO$.
So $\map {C_\OO} f$ is nonempty.
It remains to be shown that indeed:
:$\ds X_i \setminus \map {C_\OO} f = \bigcup \WW$
Consider an arbitrary $a \in X_i \setminus \map {C_\OO} f$.
Define $\ds g \in \prod_{j \mathop \le i} X_j$ by:
:$f \subseteq g$
and:
:$\map g i = a$
Then:
:$g \notin T_\OO$
and therefore there is a finite set $\PP \subseteq \OO$ such that:
:$\ds \set {x \in X: f \subseteq x, \map x i = a} = \set {x \in X: g \subseteq x} \subseteq \bigcup \PP$
and:
:$\forall V \in \PP: \set {x \in X: f \subseteq x, \map x i = a} \cap V \ne \O$
Let $\ds U = \bigcap \set {\map {\pr_i} V: V \in \PP}$.
Then $U$ is an open subset of $X_i$, $a\in U$, and:
:$\ds \set {x \in X: f \subseteq x, \map x i \in U} \subseteq \bigcup \PP$
Therefore:
:$U \cap \map {C_\OO} f = \O$
and $a \in U \in \WW$.
=== Step 2 ===
Every branch of $T_\OO$ has the greatest element.
 $B$ is a branch of $T_\OO$ without such a greatest element.
Let $f = \bigcup B$.
Let $i$ be the least element of $I$ that is not in the domain of any element of $B$.
Then:
:$\ds f \in \prod_{j \mathop < i} X_j$
Since $B$ has no greatest element, $f \notin B$.
Since $B$ is a maximal chain in $T_\OO$:
:$f \notin T_\OO$.
Let $\PP \subseteq \OO$ be a finite collection such that:
:$\ds \set {x \in X: f \subseteq x} \subseteq \bigcup \PP$
Let $m$ be the greatest element of the finite set:
:$\set {j \in I: j < i \text{ and } \exists V \in \PP: \paren {\map {\pr_j} V \ne X_j} }$
Let $g$ be any element of $B$ that is defined on $m$.
Consider an arbitrary $x \in X$ such that $g \subseteq x$.
Let $y \in X$ be defined by $f \subseteq y$ and $\map y j = \map x j$ for every $j \ge i$, and choose $V \in \PP$ such that $y \in V$.
Because $V$ does not "take into account" the values of $\map x j$ for $m < j < i$:
:$x \in V$
Thus:
:$\set {x \in X: g \subseteq x} \subseteq \bigcup \PP$
in contradiction of the fact that $g \in T_\OO$.
=== Step 3 ===
Every maximal element of $T_\OO$ is an element of $X$:
Because $\map {C_\OO} f \ne \O$:
:an element $f \in T_\OO \setminus X$ cannot be maximal in $T_\OO$.
=== Conclusion ===
Now it can be shown that there is $f \in X$ such that $f \notin \bigcup \OO$.
By hypothesis, $T_\OO$ has a branch $B$.
Let $f$ be the greatest element of $B$.
Then $f$ is a maximal element of $T_\OO$.
Therefore $f \in X$.
Therefore the set $\set f = \set {x \in X: f \subseteq x}$ is not covered by any finite subcollection of $\OO$.
Hence $f \notin \bigcup \OO$.
== Corollaries without Axiom of Choice ==
=== Corollary 1 ===
The Cartesian product of a finite indexed family of compact topological spaces is compact.
=== Corollary 2 ===
Let $I$ be a well-orderable set.
Let $\family {X_i}_{i \mathop \in I}$ be an indexed family of compact topological spaces.
Let the Cartesian product of all nonempty closed subsets of all $X_i$ be nonempty:
:$\ds \prod \set {C: C \ne \O \text { and } C \text { is closed in } X_i \text { for some } i \in I} \ne \O$
Then $\ds \prod_{i \mathop \in I} X_i$ is compact.
=== Proof ===
Let $<$ be a well-order relation on $I$.
Denote $\ds X = \prod_{i \mathop \in I} X_i$.
Let $F$ be the tree:
:$\ds \paren {\bigcup_{i \mathop \in I} \prod_{j \mathop < i} X_j} \cup X$
ordered by set inclusion.
To apply the theorem, it is enough to verify that:
:If $T$ is a subtree of $F$ such that:
::for every $i \in I$ and every $\ds f \in T \cap \prod_{j \mathop < i} X_j$, the set $\set {\map g i: g \in T, \ f \subsetneq g}$ is closed in $X_i$
:then $T$ has a branch.
Let:
:$\ds e \in \prod \set {C: C \ne \O \text { and } C \text{ is closed in } X_i \text { for some } i \in I}$
be a choice function.
That is:
:$\map e C \in C$ for every nonempty $C$ which is closed in some $X_i$.
Let $B_e$ be the minimal tree among all subtrees $S$ of $T$ with the property that:
:for every $i \in I$ and every $\ds f \in S \cap \prod_{j \mathop < i} X_j$:
::$\map e {\set {\map g i: g \in T, f \subsetneq g} } \in \set{\map g i: g \in S, f \subsetneq g}$
:unless:
::$\set {\map g i: g \in T, f \subsetneq g} = \O$
Such subtrees of $T$ exist because $T$ itself is one such.
The minimal such subtree is the intersection of all such subtrees.
It can be shown that $B_e$ is a branch by assuming that it is not, considering the minimal $i\in I$ where it "branches", and arriving at a contradiction with its minimality.
=== Applications ===
The proof that $\closedint 0 1^\Z$ is compact does not require the Axiom of Choice, because the product of all nonempty closed subsets of $\closedint 0 1$ contains, for example, the greatest lower bound function $\inf$ (restricted to the collection of closed subsets of $\closedint 0 1$):
:$\ds \set {\struct {C, \inf C} : C \ne \O \text { and } C \text{ is closed in } \closedint 0 1} \in \prod \set {C: C \ne \O \text { and } C \text{ is closed in } \closedint 0 1}$
(Here $\ds \set {\struct {C, \inf C}: C \ne \O \text { and } C \text{ is closed in } \closedint 0 1}$ is a way to write a function $f$ defined on the collection of all nonempty closed subsets of $\closedint 0 1$ by $f \sqbrk C = \inf C$.)
== Also see ==
* Mappings of Initial Intervals of a Well-Ordered Set Ordered by Inclusion: such mappings form a set-theoretic tree.
Category:Compact Spaces
Category:Product Spaces