# 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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$\copyright \ \ 2012$