# countable product of $\omega_1 + 1$

How can I prove that the topological countable product of $\omega_1 + 1$ (with the order topology) is the union of $\omega_1$ closed nowhere dense sets?

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The first thing is to figure out how to recognize a nowhere dense set. Consider a basic open set in the product. Without loss of generality it has the form $$B_0\times\dots\times B_n\times(\omega_1+1)^\omega\;,$$ where $B_0,\dots,B_n$ are open sets in $\omega_1+1$. Every non-empty open set in the product contains such a set, so if $U$ is open in $(\omega_1+1)^\omega$, there must be infinitely many factors on which its projection is the whole factor. Thus, if a closed set has bounded projections on all but finitely many factors, it must be closed and nowhere dense.
Now for $\alpha<\omega_1$ let $$F_\alpha=\{x\in(\omega_1+1)^\omega:\forall n\in\omega(x_n\le\alpha)\}\;;$$ you shouldn’t have any trouble verifying that the $F_\alpha$ are closed and nowhere dense. They cover everything except the points that are $\omega_1$ on one or more factors. What if for $n\in\omega$ you now let $$H_n=\{x\in(\omega_1+1)^\omega:x_n=\omega_1\}\;?$$
@ana: That’s what the first paragraph is about. Look at an $F_\alpha$, for instance: can it contain any basic open set? No, because its projection on every factor is $[0,\alpha]$, and every basic open set has projection $[0,\omega_1]$ on all but finitely many factors. Thus, $F_\alpha$ is in a sense too small to contain a basic open set. Of course that means that it can’t contain any non-empty open set, and since it’s already closed, it must be nowhere dense. – Brian M. Scott Mar 24 '12 at 0:30
@ana: The argument for $H_n$ is similar: it has projection $[0,\omega_1]$ on only one factor, so it also can’t contain a non-empty open set. – Brian M. Scott Mar 24 '12 at 0:30