Here is a proof of the Tychonoff theorem that relies on the well-ordering theorem rather than on Zorn’s lemma. First, prove the following version of the tube lemma; then prove the theorem.
Lemma. Let \(\mathcal{A}\) be a collection of basis elements for the topology of the product space \(X \times Y\), such that no finite subcollection of \(\mathcal{A}\) covers \(X \times Y\). If \(X\) is compact, there is a point \(x \in X\) such that no finite subcollection of \(\mathcal{A}\) covers the slice \(\ts{x} \times Y\).
Theorem. An arbitrary product of compact spaces is compact in the product topology.
Proof. Let \(\ts{X_\alpha}_{\alpha \in J}\) be an indexed family of compact spaces, let
\[ X = \prod_{\alpha \in J} X_\alpha. \]
Let \(\pi_\alpha: X \to X_\alpha\) be the projection map. Well-order \(J\), once and for all, in such a way that \(J\) has a largest element.
- Let \(\beta \in J\). Suppose points \(p_i \in X_i\) are given, for all \(i < \beta\). For any \(\alpha < \beta\), let \(Y_\alpha\) denote the subspace of \(X\) defined by the equation
\[ Y_\alpha = \ts{\mathbf{x} \mid \pi_i(\mathbf{x}) = p_i \text{ for } i \leq \alpha}. \]
Note that if \(\alpha < \alpha'\), then \(Y_\alpha \supset Y_{\alpha'}\). Show that if \(\mathcal{A}\) is a finite collection of basis elements for \(X\) that covers the space
\[ Z_\beta = \bigcap_{\alpha < \beta} Y_\alpha = \ts{\mathbf{x} \mid \pi_i(\mathbf{x}) = p_i \text{ for } i < \beta}, \]
then \(\mathcal{A}\) actually covers \(Y_\alpha\) for some \(\alpha < \beta\). [Hint: If \(\beta\) has an immediate predecessor in \(J\), let \(\alpha\) be that immediate predecessor. Otherwise, for each \(A \in \mathcal{A}\), let \(J_A\) denote the set of those indices \(i < \beta\) for which \(\pi_i(A) \neq X_i\); the union of the sets \(J_A\), for \(A \in \mathcal{A}\), is finite; let \(\alpha\) be the largest element of this union.]
- Assume \(\mathcal{A}\) is a collection of basis elements for \(X\) such that no finite subcollection of \(\mathcal{A}\) covers \(X\). Show that one can choose points \(p_i \in X_i\) for all \(i\), such that for each \(\alpha\), the space \(Y_\alpha\) defined in (a) cannot be finitely covered by \(\mathcal{A}\). When \(\alpha\) is the largest element of \(J\), one has a contradiction. [Hint: If \(\alpha\) is the smallest element of \(J\), use the preceding lemma to choose \(p_\alpha\). If \(p_i\) is defined for all \(i < \beta\), note that (a) implies that the space \(Z_\beta\) cannot be finitely covered by \(\mathcal{A}\) and use the lemma to find \(p_\beta\).]
Solution
Goal: Prove the Lemma (generalized Tube Lemma for slices) and complete the transfinite induction proof of the Tychonoff Product Theorem via the Well-Ordering Theorem.
1. Proof of the Lemma (Tube Lemma version for slices):Proof:
1.1. Suppose for contradiction that for every \(x \in X\), there exists a finite subcollection \(\mathcal{A}_x \subset \mathcal{A}\) that covers \(\{x\} \times Y\).
1.2. The union \(U_x = \bigcup_{A \in \mathcal{A}_x} A\) is an open set in \(X \times Y\) containing the slice \(\{x\} \times Y\).
1.3. Because \(Y\) is compact (or simply by the Tube Lemma since \(\{x\}\) is compact and \(U_x\) contains the slice), there exists an open neighborhood \(W_x\) of \(x\) in \(X\) such that \(W_x \times Y \subseteq U_x\).
1.4. The collection \(\{W_x\}_{x \in X}\) forms an open covering of the compact space \(X\).
1.5. There exists a finite subcover \(\{W_{x_1}, \dots, W_{x_k}\}\) of \(X\).
1.6. Then the finite collection \(\bigcup_{j=1}^k \mathcal{A}_{x_j} \subset \mathcal{A}\) covers \(\bigcup_{j=1}^k (W_{x_j} \times Y) = X \times Y\), contradicting the hypothesis that no finite subcollection of \(\mathcal{A}\) covers \(X \times Y\).
1.7. Thus there exists \(x \in X\) such that no finite subcollection of \(\mathcal{A}\) covers \(\{x\} \times Y\).
2. Proof of Part (a):Proof:
2.1. Let \(\mathcal{A}_0 \subset \mathcal{A}\) be a finite collection of basic open sets covering \(Z_\beta\).
2.2. For each \(A \in \mathcal{A}_0\), let \(J_A = \{i < \beta \mid \pi_i(A) \neq X_i\}\). Since each \(A\) is a basic open set in the product topology, \(J_A\) is finite.
2.3. Because \(\mathcal{A}_0\) is finite, the union \(F = \bigcup_{A \in \mathcal{A}_0} J_A \subset \{i \in J \mid i < \beta\}\) is finite.
2.4. If \(F = \varnothing\), let \(\alpha\) be any index less than \(\beta\). If \(F \neq \varnothing\), let \(\alpha = \max F < \beta\).
2.5. Let \(\mathbf{x} \in Y_\alpha\), so \(\pi_i(\mathbf{x}) = p_i\) for all \(i \le \alpha\).
2.6. Construct a point \(\mathbf{x}' \in Z_\beta\) by setting \(\pi_i(\mathbf{x}') = p_i\) for all \(i < \beta\), and \(\pi_i(\mathbf{x}') = \pi_i(\mathbf{x})\) for all \(i \ge \beta\).
2.7. Since \(\mathcal{A}_0\) covers \(Z_\beta\), \(\mathbf{x}' \in A\) for some \(A \in \mathcal{A}_0\).
2.8. For all \(i \in J\), if \(i \le \alpha\), \(\pi_i(\mathbf{x}) = p_i = \pi_i(\mathbf{x}') \in \pi_i(A)\). If \(\alpha < i < \beta\), \(i \notin F\), so \(\pi_i(A) = X_i\), hence \(\pi_i(\mathbf{x}) \in \pi_i(A)\). If \(i \ge \beta\), \(\pi_i(\mathbf{x}) = \pi_i(\mathbf{x}') \in \pi_i(A)\).
2.9. Thus \(\mathbf{x} \in A\), so \(\mathcal{A}_0\) covers \(Y_\alpha\).
3. Proof of Part (b) and Theorem:Proof:
3.1. Assume \(\mathcal{A}\) is a basic open cover of \(X\) with no finite subcover.
3.2. Well-order \(J\) so that \(J\) has a maximum element \(\Omega = \max J\).
3.3. By transfinite induction, we choose \(p_\beta \in X_\beta\) for each \(\beta \in J\) such that \(Y_\beta\) cannot be covered by any finite subcollection of \(\mathcal{A}\):- For the minimal element \(0 \in J\), \(X \cong X_0 \times \prod_{\alpha > 0} X_\alpha\). By 1, choose \(p_0 \in X_0\) such that \(Y_0 = \{p_0\} \times \prod_{\alpha > 0} X_\alpha\) is not finitely covered.- Assuming \(p_i\) is defined for all \(i < \beta\), Part (a) implies \(Z_\beta\) cannot be finitely covered by \(\mathcal{A}\).- Writing \(Z_\beta \cong X_\beta \times \prod_{j > \beta} X_j\), applying 1 to the compact factor \(X_\beta\) produces \(p_\beta \in X_\beta\) such that \(Y_\beta\) cannot be finitely covered by \(\mathcal{A}\).
3.4. For the maximal element \(\Omega = \max J\), \(Y_\Omega = \{(p_i)_{i \in J}\}\) is a single point in \(X\).
3.5. Since \(\mathcal{A}\) covers \(X\), the single point \(Y_\Omega\) is covered by a single element of \(\mathcal{A}\), contradicting the fact that \(Y_\Omega\) cannot be finitely covered.
3.6. Therefore, every basic open cover of \(X\) has a finite subcover, proving \(X = \prod_{\alpha \in J} X_\alpha\) is compact. Q.E.D.