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Set $\Bbb T^n = \Bbb R^n \setminus \Bbb Z^n$. Define $d : \Bbb T^n \times \Bbb T^n \to \Bbb R$ by

$$d(x+\Bbb Z^n,y+\Bbb Z^n)=\inf\{\|v-w\|:v\in x+\Bbb Z^n, \text{ and } w \in y+ \Bbb Z^n\}$$

Show that $ \Bbb T^n$ is complete and compact

My thought : I can prove $\Bbb T^n$ is complete by using completeness of $\Bbb R^n$. When it comes to compactness, I tried to give a homeomorphism $\psi : S^1\times \cdots \times S^1 \to \Bbb T^n$ by $\psi((e^{2 \pi ix_1}, e^{2 \pi ix_2},\ldots, e^{2 \pi ix_n}))= x+\Bbb Z$, where $x=(x_1,x_2,\ldots,x_n).$ Clealy, $\psi$ is bijective and If we can show it is continuous, $\psi$ is homeomorphism because $S^1\times \cdots \times S^1$ is compact and $\Bbb T^n$ is hausdorff. Thus, $\Bbb T^n$ is compact

However, I was stuck in showing $\psi$ is continuous.While posting on it, I realized it suffices to show $\Bbb T^n$ is bounded and closed. Anyway, Could you give me a few hints..??

Michael Hardy
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fivestar
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    Show that there is a surjective function $[0,1]^n \rightarrow \mathbb T^n$ and try to prove that for topological spaces the image of a compact set under a continuous function is compact. – Noel Lundström Mar 03 '20 at 15:07
  • I edited this question to change $\mathbb R^n \backslash \mathbb Z^n$ to $\mathbb R^n \setminus \mathbb Z^n$ and $||v-w||$ to $|v-w|.$ If the reason why that last is standard usage is not conspicuous enough, contrast $||a|| ||b||$ with $|a||b|. \qquad$ – Michael Hardy Mar 03 '20 at 15:09
  • @NoelLundström Thank you for a hint. So, I tried to gave a map $x$ into $x+\Bbb Z^n$. But I was stuck in showing this map is continuous. The thing is that it is difficult for me to describe the Ball with metric in $\Bbb T^n$. Could you give me a more hint to show it is continuous? – fivestar Mar 04 '20 at 03:36
  • @MichaelHardy Thank you for editting – fivestar Mar 04 '20 at 03:36
  • @fivestar The map in question is $[0,1]^n \rightarrow \mathbb R ^n \rightarrow \mathbb R^n / \mathbb Z^n$. Show that each of the maps are continuous and thus their composition is aswell. – Noel Lundström Mar 04 '20 at 07:20
  • @NoelLundström I am sorry :( I can't find those map from $[0,1]^n$ onto $\Bbb R^n$ – fivestar Mar 04 '20 at 08:16
  • It's the inclusion map. You take a point $(x_1,...,x_n)$ in $[0,1]^n$ to $(x_1,...,x_n)$ in $\mathbb R ^ n$ – Noel Lundström Mar 04 '20 at 11:12
  • @NoelLundström And then, give a map, $f : \Bbb R^n \rightarrow \Bbb R^n/\Bbb Z$ by $f(x)=x+\Bbb Z^n$. However, How to show $f$ is continuous..? I know surjective function induce Quotient topology, but in this case I have to show $f$ is continuous based on the metric of codomain – fivestar Mar 04 '20 at 11:36
  • @fivestar Ah yes that does make the problem more difficult. I will think about it! – Noel Lundström Mar 04 '20 at 21:13

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