Notes

Description:

Form 43 implies Form 113

Content: In this note we prove that Form 43 (Dependent Choice) implies Form 113 (A countable product of compact topological spaces is compact.)

Proof: Our proof is based on the proof of theorem 6.1 in Good/Tree [1995]. Let \(\{X_\alpha: \alpha < \omega\}\) be a countable set of compact topological spaces.  Let \(X = \prod_{\alpha\in\omega} X_\alpha\) and for \(\beta < \omega\) define \(p_\beta : X\to\prod_{\alpha\le\beta}X_\alpha\)by \(p_\beta(x_0,x_1,\ldots) = (x_0,x_1,\ldots,x_\beta)\).To show that \(X\) is compact let \(\cal C\) be a collection of closed sets in \(X\) with the finite intersection property.  We will use Dependent Choice to choose a sequence \((x_0,x_1,\ldots)\)with the property that for all \(\alpha <\omega\),\((x_0,x_1,\ldots,x_\alpha)\in\bigcap\{\overline{p_\alpha(C)}:C \in \cal C\}\).  Then we will prove, as in Good/Tree [1995], that \((x_1,x_2,\ldots)\in\bigcap\cal C\).

Definition: If \(x_0\in X_0\) and \(\alpha < \kappa\) we say \(x_0\) can be extended to level \(\alpha\) if there is a sequence \((x_1,\ldots,x_\alpha)\) such that \((x_0,x_1,\ldots,x_\alpha)\in\bigcap\{\overline{p_\alpha(C)} : C\in\cal C\}\). Similarly,if \(\alpha < \beta < \kappa\) and \((x_0,x_1,\ldots,x_\alpha) \in\bigcap\{\overline{p_\alpha(C)} : C\in\cal C\}\) then we say \((x_0,x_1,\ldots,x_\alpha)\) can be extended to level \(\beta\) if there is a sequence \((y_{\alpha+1},\ldots,y_\beta)\) such that \((x_0,x_1,\ldots,x_\alpha,y_{\alpha+1},\ldots,y_\beta)\in\bigcap\{\overline{p_\beta(C)} : C\in\cal C\}\).

Definition: If \(\beta \le \gamma < \kappa\) define \[p_{\beta,\gamma}:\prod_{\alpha\le\gamma} X_\alpha \to\prod_{\alpha\le\beta} X_\alpha\] by \[p_{\beta,\gamma}(x_0,x_1,\ldots,x_\beta,\ldots,x_\gamma) =(x_0,x_1,\ldots,x_\beta).\]

For every \(\alpha < \kappa\), let \(X_{0,\alpha} = \{ x\in X_0 : x\)can be extended to level \(\alpha \}\).  \(X_{0,\alpha}\) is closed in \(X_0\) because \(X_{0,\alpha} =p_{0,\alpha}(\bigcap\{\overline{p_\alpha(C)} : C\in\cal C\})\) and

  1. \(\prod_{0\le\beta\le\alpha} X_\beta =X_0 \times \prod_{1\le\beta\le\alpha} X_\beta\) is compact.
  2. \(\bigcap\{\overline{p_\alpha(C)} : C\in\cal C\}\) is closed in \(\prod_{0\le\beta\le\alpha} X_\beta\).
  3. If \(Y\) and \(Z\) are compact topological spaces and\(D\) is closed in \(Y\times Z\) then the projection of \(D\) onto\(Y\) is closed.
We also note that \(X_{0,\alpha} \ne\emptyset\) since \(\bigcap\{\overline{p_\alpha(C)} : C\in\cal C\}\) is non-empty (by the compactness of \(\prod_{0\le\beta\le\alpha} X_\beta\)).  We can conclude that the family \(\{ X_{0,\alpha} : \alpha <\kappa \}\) has the finite intersection property since \(X_{0,\alpha_1}\cap\ldots\cap X_{0,\alpha_k} =X_{0,\alpha}\) where \(\alpha = \max\{\alpha_1,\ldots,\alpha_k\}\). Since \(X_0\) is compact \(\bigcap_{\alpha<\kappa} X_{0,\alpha} \ne\emptyset\). Choose \(x_0\in \bigcap_{\alpha<\kappa} X_{0,\alpha}\).

Assume that \(x_0,\ldots,x_\beta\) have been chosen so that\((x_0,\ldots,x_\beta)\) is extendible to level \(\alpha\) for every \(\alpha\) such that \(\beta < \alpha < \kappa\). For every such \(\alpha\), let \(X_{\beta +1,\alpha} = \{x\in X_{\beta+1} :(x_0,\ldots,x_\beta,x) \) is extendible to level \(\alpha\}\).

Claim:  \(X_{\beta+1,\alpha}\) is closed.

First note that \[X_{\beta+1,\alpha} = \ x\in X_{\beta+1} : (x_0,\ldots,x_\beta,x) \in p_{\beta+1,\alpha}(\bigcap\{\overline{p_\alpha(C)} : C\in\cal C\}).\] \(\bigcap\{ \overline{p_\alpha(C)} : C\in\cal C\}\) is closed in \(\prod_{\gamma\le\alpha} X_\gamma = \left(\prod_{\gamma\le\beta+1} X_\gamma\right)\times\left(\prod_{\beta+1<\gamma\le\alpha} X_\gamma\right)\). Therefore, by C. above, \(p_{\beta+1,\alpha}(\bigcap\{\overline{p_\alpha(C)}:C\in\cal C\})\) is closed.  Assume \(y\in X_{\beta+1}\) is in the closure of \(X_{\beta+1,\alpha}\).  We will complete the proof of the claim by showing that \(y\in X_{\beta+1,\alpha}\). Let \(N\) be aneighborhood of \((x_0,\ldots,x_\beta,y)\) in\(\prod_{\gamma\le\beta+1} X_\gamma\). Then the projection of \(N\) onto \(X_{\beta+1}\) is a neighborhood of \(y\).  Therefore there is an \(x\in X_{\beta+1,\alpha}\) such that \(x\) is in this projection and therefore, the sequence \((x_0,\ldots,x_\beta,x)\) is in \(N\).  Also \((x_0,\ldots,x_\beta,x)\) is in \(p_{\beta+1,\alpha}(\bigcap\{\overline{p_\alpha(C)} : C\in\cal C\})\). This shows that \((x_0,\ldots,x_\beta,y)\) is in the closure of \(p_{\beta+1,\alpha}(\bigcap\{\overline{p_\alpha(C)} : C\in\cal C\})\) and hence \((x_0,\dots,x_\beta,y)\) is in \(p_{\beta+1,\alpha}(\bigcap\{\overline{p_\alpha(C)} : C\in\cal C\})\).\(y\) is therefore in \(X_{\beta+1,\alpha}\). As above, the family\(\{ X_{\beta+1,\alpha} : \beta < \alpha < \kappa\}\) has the finite intersection property. Since \(X_{\beta+1}\) is compact, \(\bigcap_{\beta<\alpha<\kappa} X_{\beta+1,\alpha} \ne\emptyset\). Choose \(x_{\beta+1} \in \bigcap_{\beta<\alpha<\kappa} X_{\beta+1,\alpha}\).

The argument that \((x_0,x_1,\ldots)\) is in \(\bigcap\cal C\) is by contradiction.  Suppose the assertion is false. Then there is an open set \(U = \prod_{\alpha<\omega} U_\alpha\) and \(C\in\cal C\) such that \((x_0,x_1,\ldots)\in U\) and \(U\cap C = \emptyset\). Choose\(\gamma\in\omega\) so that \(U_\alpha = X_\alpha\) whenever \(\alpha >\gamma\), and let \(V = \prod_{\alpha\le\gamma} U_\alpha\).  \(V\) is open in\(\prod_{\alpha\le\gamma}X_\alpha\).  Further \((x_0,x_1,\ldots,x_\gamma)\in V\) hence \(V\cap p_\gamma(C) \ne \emptyset\).  This means that there is some sequence \((y_0,y_1,\ldots)\in C\) with \((y_0,y_1,\ldots,y_\gamma)\in V\).  Which (by definition of \(V\)) implies that \((y_0,y_1,\ldots)\in U\). This contradicts \(U\cap C=\emptyset\).

Howard-Rubin number: 7

Type: Proof

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