Consider the topological space $(\mathbb{Z},\mathcal{P}(\mathbb{Z}))$. It is clearly Hausdorff since $(\mathbb{Z},\mathcal{P}(\mathbb{Z}))$ can be seen as a metric space where the topology is induced by the discrete metric. Furthermore, it is clearly second-countable since a countable basis is given by $$\{\{n\} : n \in \mathbb{Z}\}$$ Now my question is, is this space also a manifold and of which dimension? I would say, that it is a $0$-dimensional manifold since if $x \in \mathbb{Z}$, $\{x\}$ is a neighbourhood of $x$ and clearly homeomorphic to the one-point set $\mathbb{R}^0$. Is that correct?
Is $(\mathbb{Z},\mathcal{P}(\mathbb{Z}))$ a manifold?
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general-topology
manifolds
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2I believe this provides an answer: http://math.stackexchange.com/questions/1031121/build-a-topological-manifold-starting-from-a-set – 2017-02-09
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1Correct. A (countable) discrete space is a manifold of dimension $0$. (Depending on the definition of manifold in use, we may not need to restrict to countable discrete spaces.) – 2017-02-09
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0@DanielFischer Thank you. I think the problem was that I am not entirely sure what $\mathbb{R}^0$ is. I know it is a one point set and abstractly does only contain the empty function. Is it also correct that we can say $\mathbb{R}^0=\{0\}$? – 2017-02-09
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2Yes, $\mathbb{R}^0 = \{0\}$ is a typical representation. The point is also often called $\ast$. – 2017-02-09