Tychonoff's Theorem Without Choice
![]() | This article needs proofreading. Please check it for mathematical errors. If you believe there are none, please remove {{Proofread}} from the code.To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Proofread}} from the code. |
![]() | This page has been identified as a candidate for refactoring of advanced complexity. In particular: This should not be specified as a separate result from Tychonoff's Theorem -- the statement of the result is the same, it's just its execution which is different. Hence it needs to be transcluded into the main body of Tychonoff's Theorem as a Proof 2. Corollaries to be extracted and disposed as appropriate. Lose the "preliminaries", as this does not fit $\mathsf{Pr} \infty \mathsf{fWiki}$ style. Until this has been finished, please leave {{Refactor}} in the code.
New contributors: Refactoring is a task which is expected to be undertaken by experienced editors only. Because of the underlying complexity of the work needed, it is recommended that you do not embark on a refactoring task until you have become familiar with the structural nature of pages of $\mathsf{Pr} \infty \mathsf{fWiki}$.To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Refactor}} from the code. |
![]() | This article needs to be linked to other articles. In particular: Various places You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{MissingLinks}} from the code. |
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 $\ds f \in T_\OO \cap \prod_{j \mathop < i} X_j$:
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.
![]() | It has been suggested that this page or section be merged into Topological Product of Compact Spaces. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Mergeto}} from the code. |
$\blacksquare$
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.
![]() | This article, or a section of it, needs explaining. In particular: "It can be shown that $B_e$ is a branch": so this needs to be done. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
$\blacksquare$
Applications
![]() | This page or section has statements made on it that ought to be extracted and proved in a Theorem page. In particular: Find a way to express this as a result in its own right, and link to it perhaps as an "also see". You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by creating any appropriate Theorem pages that may be needed. To discuss this page in more detail, feel free to use the talk page. |
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 $f \sqbrk C = \inf C$.)
![]() | This article, or a section of it, needs explaining. In particular: Why does "the product of all nonempty closed subsets of $\closedint 0 1$ contains the greatest lower bound function $\inf$" not require the Axiom of Choice? This section needs to be removed from here and placed into its own properly-structured proof page. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
Also see
- Mappings of Initial Intervals of a Well-Ordered Set Ordered by Inclusion: such mappings form a set-theoretic tree.