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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 $H_n=\{x\in(\omega_1+1)^\omega:x_n=\omega_1\}\;?$

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    Thank you very much Brian, now I see everything more clear, you really helped me a lot :)2012-03-26