Looking for another closed and discrete subspace of a product space

Let \omega_1 be the first uncountable ordinal. In a previous post called Looking for a closed and discrete subspace of a product space, it was shown that the product space \mathbb{R}^c, the product of continuum many copies of the real line \mathbb{R}, contains a closed and discrete subset of cardinality continuum. This example shows that a product space of uncountably many copies of a “nice” space is “big and wide” enough to hide uncountable closed and discrete sets even when the product space is separable. This post reinforces this same fact by showing that \mathbb{R}^{\omega_1} contains a closed and discrete subset of cardinality \omega_1. It follows that for any uncountable cardinal \tau, the product space \mathbb{R}^\tau contains an uncountable closed and discrete subset, i.e., the product of uncountably many copies of the real line \mathbb{R} has uncountable extent.

Let \omega be the first infinite ordinal, i.e., the set of all nonnegative integers. Consider \omega^{\omega_1}, the product of \omega_1 many copies of \omega with the discrete topology. Since \omega^{\omega_1} is a closed subspace of \mathbb{R}^{\omega_1}, it suffices to show that \omega^{\omega_1} has an uncountable closed and discrete subset.

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The Construction

We now construct an uncountable closed and discrete subset of \omega^{\omega_1}. Let \delta be an infinite ordinal such that \omega<\delta<\omega_1. Let W=\left\{\alpha: \delta \le \alpha<\omega_1 \right\}. For each \alpha \in W, let Y_\alpha=\left\{\beta<\omega_1: \beta<\alpha \right\}. We can also use interval notations: W=[\delta,\omega_1) and Y_\alpha=[0,\alpha). Consider Y_\alpha as a space with the discrete topology. Then it is clear that \omega^{\omega_1} is homeomorphic to the product space \prod_{\alpha \in W} Y_\alpha. Thus the focus is now on finding an uncountable closed and discrete subset of \prod_{\alpha \in W} Y_\alpha.

One interesting fact about the space \prod_{\alpha \in W} Y_\alpha is that every function f \in \prod_{\alpha \in W} Y_\alpha is a pressing down function. That is, for every f \in \prod_{\alpha \in W} Y_\alpha, f(\alpha)<\alpha for all \alpha \in W. Note that f is defined on W, a closed and unbounded subset of \omega_1 (hence a stationary set). It follows that for each f \in \prod_{\alpha \in W} Y_\alpha, there is a stationary set S \subset W and there exists \rho<\omega_1 such that f(\alpha)=\rho for all \alpha \in S. This fact is called the pressing down lemma and will be used below. See this post for more information about the pressing down lemma.

For each \gamma \in W, let h_\gamma: Y_{\gamma+1} \rightarrow \delta be a one-to-one function. For each \gamma \in W, define t_\gamma \in \prod_{\alpha \in W} Y_\alpha as follows:

    t_\gamma(\alpha) = \begin{cases} h_\gamma(\alpha) & \mbox{if } \delta \le \alpha \le \gamma \\ \gamma & \mbox{if } \gamma < \alpha <\omega_1  \end{cases}

Note that each t_\gamma is a pressing down function. Thus each t_\gamma \in \prod_{\alpha \in W} Y_\alpha. Let T=\left\{t_\gamma: \gamma \in W \right\}. Clearly t_\gamma \ne t_\mu if \gamma \ne \mu. Thus T has cardinality \omega_1. We claim that T is a closed and discrete subset of \prod_{\alpha \in W} Y_\alpha. It suffices to show that for each f \in \prod_{\alpha \in W} Y_\alpha, there exists an open set V with f \in V such that V contains at most one t_\gamma.

Let f \in \prod_{\alpha \in W} Y_\alpha. As discussed above, there is a stationary set S \subset W and there exists \rho<\omega_1 such that f(\alpha)=\rho for all \alpha \in S. In particular, choose \mu, \lambda \in S such that \mu \ne \lambda. Thus f(\mu)=f(\lambda)=\rho. Let V be the open set defined by:

    V=\left\{g \in \prod_{\alpha \in W} Y_\alpha: g(\mu)=g(\lambda)=\rho \right\}

Clearly, f \in V. We show that if t_\gamma \in V, then \gamma=\rho. Suppose t_\gamma \in V. Then t_\gamma(\mu)=t_\gamma(\lambda)=\rho. Consider two cases: Case 1: \delta \le \mu, \lambda \le \gamma; Case 2: one of \mu and \lambda>\gamma. The definition of t_\gamma indicates that t_\gamma=h_\gamma on the interval [\delta, \gamma]. Note that h_\gamma is a one-to-one function. Since t_\gamma(\mu)=t_\gamma(\lambda), it cannot be that \mu, \lambda \in [\delta, \gamma], i.e., Case 1 is not possible. Thus Case 2 holds, say \mu>\gamma. Then by definition, t_\gamma(\mu)=\gamma. Putting everything together, \gamma=t_\gamma(\mu)=t_\gamma(\lambda)=\rho. Thus V \cap T \subset \left\{t_\rho \right\}. This concludes the proof that the set T is a closed and discrete subset of \prod_{\alpha \in W} Y_\alpha. \blacksquare

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\copyright \ 2014 \text{ by Dan Ma}

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