Timeline for A kind of orthogonal subtorus

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Mar 6, 2012 at 13:32	comment	added	user21706		Maybe I have concluded: if $k = (k_1, \ldots, k_n) \in \mathbb{Z}^n$ with $\gcd(k_1, k_2, \ldots, k_n) = 1$ then for the generalized Bezout's lemma exists $m \in \mathbb{Z}^n$ such that $k \cdot m = 1$, then for any $x \in \mathbb{R}^n$ such that $k \cdot x = n \in \mathbb{Z}$ we have $k \cdot (x + m) = n + 1$ and $P(x) = P(x + m) \in S_n \cap S_{n+1}$. Can someone check my reasoning?
Mar 6, 2012 at 13:10	comment	added	user21706		I thought that if $P : \mathbb{R}^n \to \mathbb{T}^n$ is the standard projection then $S=\bigcap_{n \in \mathbb{Z}} S_n$ , where $S_n := P(\{x \in \mathbb{R}^n : k \cdot x = n\})$ . Now $\{x \in \mathbb{R}^n : k \cdot x = n\}$ is connected because is an affine hyperplane, then $S_n$ is connected because is the continuous image of a connected set. Remains to prove something like $\bigcap_{n \in \mathbb{Z}} S_n \neq \emptyset$
Mar 6, 2012 at 12:33	comment	added	user21706		Well, so I can argue as follows: $S$ is closed because is the pre-image of the closed set $\{0_{\mathbb{T}^n}\}$ by the continuous map $\mathbb{T}^n \to \mathbb{T} : x \mapsto k \cdot x$ , then $S$ is compact because $\mathbb{T}^n$ is compact. But how I can show that $S$ is connected when $k \neq ih$ for all $i =2,3,\ldots$ and $h \in \mathbb{Z}^d$?
Mar 6, 2012 at 11:10	history	answered	James Cranch	CC BY-SA 3.0