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Here is the Well known proposition. I assumed that K is the compact subset of ω^ω and it contains an open set Ns, so I thought if K is compact so Ns is because Ns is clopen set, If I can not find a finitely many subcover of Ns, solution is done. But I could not construct and don’t know how to start

Etü
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1 Answers1

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Let $K$ be compact in $\omega^\omega$. Suppose $U$ is open and non-empty (so pick $x \in U$) where $U \subseteq K$.

Then there is a basic open subset $\prod_{n \in \omega} U_n$ (so we have a finite subset $F \subseteq \omega$ such that $n \notin F \to U_n = \omega$) such that $x \in \prod_{n \in \omega} U_n \subseteq U$.

If $m \notin F$ we then have $U_m = \omega \subseteq \pi_m[U] \subseteq \pi_m[K]\subseteq \omega$ so that $\pi_m[K] = \omega$ which is not compact (infinite and discrete), while the projection $\pi_m$ is continuous and $K$ is compact. Contradiction, so no such non-empty open subset of $K$ exists.

Henno Brandsma
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