Tychonoff's Theorem Without Choice

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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.

The 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.

The Hausdorff 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.

Aiming for a contradiction, suppose $\map {C_\OO} f$ 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 $a \in U \in \WW$.

$\Box$


Step 2

Every branch of $T_\OO$ has the greatest element.

Aiming for a contradiction, suppose $B$ 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 $g \in T_\OO$.

$\Box$


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 $T_\OO$.

$\Box$


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 $f \notin \bigcup \OO$.

$\blacksquare$


Corollaries without Axiom of Choice

Corollary 1

The Cartesian product of a finite indexed family of compact topological spaces is compact.

$\blacksquare$


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.



$\blacksquare$


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