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Here is the well-known proposition, I assumed that $K$ is compact subset of $\omega^\omega$ and it contains an open Set $N_s$, so $N_s$ is clopen and thus it is compact. If I can show $N_s$ can not be covered by a finite subcover, the solution is done, but I don’t know how to start, can you please help me?

Martin Sleziak
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norito's
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2 Answers2

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It helps to use a basis. Basic open sets in Baire Space look like: $$\mathcal{U} = \{m_1\} \times \{m_2\} \times \ldots \times \{m_n\} \times \Bbb{N} \times \Bbb{N} \times \ldots \subseteq \omega^\omega,$$ for some finite sequence $m_1, \ldots, m_n \in \Bbb{N}$. If $K$ has non-empty interior, then it must contain a set of this form. These sets are indeed clopen, so by your logic, we just need to show $\mathcal{U}$ is not compact. We will find an open cover, without a finite subcover.

This is not too hard to do. Essentially, we just take open sets $$\mathcal{U}_k = \{m_1\} \times \{m_2\} \times \ldots \times\{m_n\} \times \{k\} \times \Bbb{N} \times \Bbb{N} \times \ldots,$$ where $k \in \Bbb{N}$.

Points in $\mathcal{U}$ are precisely the natural sequences which begin with $m_1, \ldots, m_n$. If we take such a sequence $(x_k) \in \mathcal{U}$, then $x_{n+1}$ must have some value, in which case, $(x_k) \in \mathcal{U}_{x_{n+1}}$. That is, $$\mathcal{U} \subseteq \bigcup_{k \in \Bbb{N}} \mathcal{U}_k.$$ That is, we have an (infinite) open cover. Does it have a finite subcover? No; in fact it doesn't even have a proper subcover. If we remove $\mathcal{U}_k$ from the union, then suddenly we are not covering the sequence $$(m_1, m_2, \ldots, m_n, k, k, k, \ldots) \in \mathcal{U}.$$ So, every single $\mathcal{U}_k$ is necessary to cover $\mathcal{U}$, and hence there cannot be a finite subcover.

Theo Bendit
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  • Thank you so much sir, what you mean x (n+1) Must have some value? – norito's Feb 11 '21 at 19:21
  • I mean that $x_{n+1}$ is some number in $\Bbb{N}$. I've made a small edit that might make it clearer, but what I'm saying is, if we let $a = x_{n+1}$, then the sequence $(x_n) \in \mathcal{U}$ must belong to $\mathcal{U}_a$. – Theo Bendit Feb 12 '21 at 02:27
  • Appreciated, thank you so much – norito's Feb 12 '21 at 05:57
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Suppose $K$ is compact. It follows that all $p_n[K]$ are also compact in $\omega$, which is a discrete space so every projection of $K$ is finite ( the only kind of compact subset of a discrete space) An open subset always has infinitely many projections that are $\omega$, by the form of standard basic open sets. From this the statement follows.

Henno Brandsma
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