Topological Spaces with Caliber Omega 1

Let \omega_1 be the first uncountable ordinal. A separable space is one that has a countable dense subset. Any separable space X satisfies this property:

    For every collection \left\{U_\alpha: \alpha < \omega_1 \right\} of non-empty open subsets of X, there is an uncountable A \subset \omega_1 such that \bigcap \limits_{\alpha \in A} U_\alpha \ne \varnothing.

Any space that has this property is said to have caliber \omega_1. Spaces that have caliber \omega_1 have the countable chain condition (abbreviated by CCC, which means that there is no uncountable pairwise disjoint collection of open subsets of the space). So we have the following implications:

    \text{separable} \Longrightarrow \text{caliber } \omega_1 \Longrightarrow \text{CCC}

The chain condition “caliber \omega_1” is an interesting one. Some of the spaces that have the CCC actually have caliber \omega_1. For example, separable spaces and products of separable spaces have caliber \omega_1. Thus the product space \left\{0,1 \right\}^{\mathcal{K}} (for any uncountable cardinal \mathcal{K}) not only has the countable chain condition. It has the stronger property of having caliber \omega_1, which may make certain proof easier to do. In this post we also provide examples to show that none of the above implications is reversible.

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Delta System Lemma

In proving that product of separable spaces has caliber \omega_1, the Delta system lemma is used. A collection \mathcal{D} of sets is said to be a Delta-system (or \Delta-system) if there is a set D such that for every A,B \in \mathcal{D} with A \ne B, we have A \cap B = D. When such set D exists, it is called the root of the Delta-system \mathcal{D}. The following is the statement of Delta-system lemma.

Lemma 1 – Delta-System Lemma
For every uncountable collection \mathcal{A} of finite sets, there is an uncountable \mathcal{D} \subset \mathcal{A} such that \mathcal{D} is a \Delta-system.

The statement of Delta-system lemma presented here is a special case for a general version (see [2], page 49).

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Products of Spaces with Caliber \omega_1

Theorem 2 below shows that whenever “caliber \omega_1” is preserved by taking product with any finite number of factors, “caliber \omega_1” is preserved by taking the product of any number of factors. As a corollary, we have the result that the product of any number of separable spaces has caliber \omega_1. The proof of Theorem 2 is similar to the one stating that whenever CCC is preserved by taking product with any finite number of factors, CCC is preserved by taking the product of any number of factors (see the previous post Product of Spaces with Countable Chain Condition).

Theorem 2
Suppose that \left\{X_\alpha: \alpha \in T \right\} is a family of spaces such that \prod \limits_{\alpha \in F} X_\alpha has caliber \omega_1 for every finite F \subset T. Then \prod \limits_{\alpha \in T} X_\alpha has caliber \omega_1.

Proof
In proving the product space \prod \limits_{\alpha \in T} X_\alpha having caliber \omega_1, it suffices to work with basic open sets of the form \prod \limits_{\alpha \in T} O_\alpha where O_\alpha=X_\alpha for all but finitely many \alpha \in T. Let \mathcal{U}=\left\{U_\beta: \beta < \omega_1 \right\} be an uncountable collection of such non-empty open sets. Our plan is to find an uncountable W_0 \subset \omega_1 such that \bigcap \limits_{\beta \in W_0} U_\beta \ne \varnothing.

\text{ }

For each U_\beta=\prod \limits_{\alpha \in T} U_{\beta,\alpha}, let F_\beta \subset T be the finite set such that \alpha \in F_\beta if and only if U_{\beta, \alpha} \ne X_\alpha. Consider \mathcal{A}=\left\{F_\beta: \beta < \omega_1 \right\}. By the Delta-system lemma, there is an uncountable W \subset \omega_1 such that \mathcal{F}=\left\{F_\beta: \beta \in W \right\} is a Delta-system. Let F be the root of this Delta-system.

\text{ }

Consider the case that the root of the Delta-system is empty. Then for any F_{\beta_1} \in \mathcal{F} and F_{\beta_2} \in \mathcal{F} where F_{\beta_1} \ne F_{\beta_2}, we have F_{\beta_1} \cap F_{\beta_2}=\varnothing. For each \beta \in W, we have U_\beta=\prod \limits_{\alpha \in T} U_{\beta,\alpha} and we choose h_\beta in \prod \limits_{\alpha \in F_\beta} U_{\beta,\alpha}. Then we can define an h in \prod \limits_{\alpha \in T} X_\alpha such that h extends h_\beta for all \beta \in W. Then in this case let W_0=W and we have \bigcap \limits_{\beta \in W_0} U_\beta \ne \varnothing. So we move onto the case that F \ne \varnothing.

\text{ }

Now assume F \ne \varnothing. Let \mathcal{U}^*=\left\{U_\beta: \beta \in W \right\}. For each U_\beta \in \mathcal{U}^* where U_\beta=\prod \limits_{\alpha \in T} U_{\beta,\alpha}, let p(U_\beta) be \prod \limits_{\alpha \in F} U_{\beta,\alpha} (i.e. p is a projection map). Let \mathcal{U}^{**}=\left\{p(U_\beta): \beta \in W \right\}.

\text{ }

Consider two cases. One is that \mathcal{U}^{**} is countable. The second is that \mathcal{U}^{**} is uncountable. Suppose \mathcal{U}^{**} is countable. Then there is an uncountable W_0 \subset W such that p(U_\gamma)=p(U_\mu) for all \gamma,\mu \in W_0. Then fix \gamma \in W_0 and choose g^* \in p(U_\gamma). Then g^* is extendable to some g in \prod \limits_{\alpha \in T} X_\alpha such that g \in U_\beta for all \beta \in W_0. Thus we have \bigcap \limits_{\beta \in W_0} U_\beta \ne \varnothing.

\text{ }

Now assume \mathcal{U}^{**} is uncountable. By assumption \prod \limits_{\alpha \in F} X_\alpha has caliber \omega_1. Then there is an uncountable W_0 \subset W such that \bigcap \limits_{\beta \in W_0} p(U_\beta) \ne \varnothing. Choose g^* \in \bigcap \limits_{\beta \in W_0} p(U_\beta). As in the previous case, g^* is extendable to some g in \prod \limits_{\alpha \in T} X_\alpha such that g \in U_\beta for all \beta \in W_0. Thus we have \bigcap \limits_{\beta \in W_0} U_\beta \ne \varnothing. \blacksquare

Corollary 3
Suppose that \left\{X_\alpha: \alpha \in T \right\} is a family of separable spaces. Then \prod \limits_{\alpha \in T} X_\alpha has caliber \omega_1.

Proof
This follows directly from Theorem 2. Note that the product of finitely many separable is separable (hence has caliber \omega_1). \blacksquare

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Examples

We now show that the following implications are not reversible.

    \text{ }

    \text{separable} \Longrightarrow \text{caliber } \omega_1 \Longrightarrow \text{CCC}

    \text{ }

To get a space with caliber \omega_1 that is not separable, consider the product of \mathcal{K} many copies of \left\{0,1 \right\} where \mathcal{K} is a cardinal number greater than continuum. Since \left\{0,1 \right\}^{\mathcal{K}} is a product of separable spaces, it has caliber \omega_1 according to Corollary 3. It is well known that the product of more than continuum many separable spaces is not separable (see Product of Separable Spaces).

To get a space with the CCC that does not have caliber \omega_1. Consider the subspace S of the product space H=\left\{0,1 \right\}^{\omega_1} (\omega_1 many copies of \left\{0,1 \right\}) where S is the set of all h \in H such that h(\alpha) \ne 0 for at most countably many \alpha < \omega_1. Note that H is a space with the CCC since it is a product of separable spaces. Furthermore S is a dense subspace of H. The property of CCC is hereditary with respect to dense subsets. Thus S has the countable chain condition. Here's a discussion of the countable chain condition.

To see that S does not have caliber \omega_1, look at the collection of open sets \left\{V_\alpha: \alpha < \omega_1 \right\} where each V_\alpha=\left\{h \in S: h(\alpha)=1 \right\}. Note that each h \in S belongs to at most countably many V_\alpha. Thus for any uncountable A \subset \omega_1, \bigcap \limits_{\alpha \in A} V_\alpha = \varnothing.

The example S shows that dense subspace of a space with caliber \omega_1 need not have caliber \omega_1.

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Kunen, K., Set Theory, An Introduction to Independence Proofs, North-Holland, Amsterdam, 1980.
  3. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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\copyright \ \ 2012

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