# 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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$\copyright \ 2014 \text{ by Dan Ma}$