Is $\{0,1\}^{\mathbb{N}}$ compact in $\mathbb{N}^{\mathbb{N}}$ ? How can i proof that?
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1Which topology are you considering? – Marco Lecci Sep 05 '19 at 08:39
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product topology of the discrete metric – Sep 05 '19 at 08:40
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This space is homeomorphic to the Cantor's set which is compact. See here https://math.stackexchange.com/questions/69905/the-cantor-set-is-homeomorphic-to-infinite-product-of-0-1-with-itself-cy – Marco Lecci Sep 05 '19 at 09:06
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The topology of $\mathbb N ^{\mathbb N}$ and $\{0,1\}^{\mathbb N}$ are both same as product topologies. Tychonof's Threorem says product of compact spaces is compact. Hence the answer is YES.
Kavi Rama Murthy
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This page (listed at pi-base as "countable power of a countable discrete space") https://topology.jdabbs.com/spaces/S000100/properties says that $\mathbb{N}^{\mathbb{N}}$ is not compact. (It says that it is actually homeomorphic to the irrationals.) But Tychonoff's theorem implies ${0, 1}^{\mathbb{N}}$ is compact. So I believe you may be mistaken when you say these are the same as product topologies. Another SE question states that ${0, 1}^{\mathbb{N}}$ is homeomorphic to a cantor set, and another SE question says a cantor set minus its endpoints is homeomorphic to the irrationals. – Geoffrey Sangston Jul 10 '20 at 17:40
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I missed a better way of showing $\mathbb{N}^{\mathbb{N}}$ is not compact. If it were then a projection map from it to one of its factors $\mathbb{N}$ would preserve compactness. But infinite discrete sets are not compact. (In general, this argument shows Tychonoff's theorem goes both directions.) – Geoffrey Sangston Jul 11 '20 at 20:10
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$\{0, 1\}^{\mathbb{N}}$ is a product of closed sets of $\mathbb{N}$, and so is closed in $\mathbb{N}^{\mathbb{N}}$. https://proofwiki.org/wiki/Product_of_Closed_Sets_is_Closed
It is compact by Tychnoff's theorem, so clearly the closure is compact. We conclude it is relatively compact in $\mathbb{N}^{\mathbb{N}}$.
Geoffrey Sangston
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