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Define an equivalence relation on $\mathbb{R}^2$ by $a \times b$ ~ $c \times d$ iff $a-c$ and $b-d$ are both integers and let $T=(\mathbb{R}^2)^*$ denote the corresponding quotient space.

(a) Show that $T$ is homeomorphic to the quotient space of $[0,1] \times [0,1]$ obtained by the equivalence relation $s \times 0 $ ~ $ s \times 1$, $0 \times t$ ~ $1\times t$, for all $s,t \in [0,1]$.

What I have so far: Given a point $a \times b \in \mathbb{R}^2$, let $s = a- |a|$ and $t = t - |b|$, (where $|x|$ is the largest integer less than x). Define $f: \mathbb{R}^2 \rightarrow ([0,1] \times [0,1])^* $ by setting $f(a \times b)$ equal to the equivalence class of $s \times t$. This is a quotient map.

(b) Show T is compact.

(c) Show that the image of a line $L$ with slope $\frac{r}{s} \in \mathbb{Q}$ under the quotient map $\mathbb{R}^2 \rightarrow T$ is compact.

Any and all help will be greatly appreciated!

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I would advise to take the map in the other direction in (a). That, among other things, gives you (b) for free. –  Daniel Fischer Dec 12 '13 at 11:16
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1 Answer 1

Define $\nu:\mathbb{R}^{2}\rightarrow\left[0,1\right)^{2}$ by $\left(a,b\right)\mapsto\left(a-\lfloor a\rfloor,b-\lfloor b\rfloor\right)$ and denote the quotient topology on $\left[0,1\right)^{2}$ by $\mathcal{O}_{\nu}$. Here $\lfloor a\rfloor$ is the largest integer that does not exceed $a$.

Then $T$ and $\left(\left[0,1\right)^{2},\mathcal{O}_{\nu}\right)$ are homeomorphic.

Let $\iota:\left[0,1\right]^{2}\rightarrow\mathbb{R}^{2}$ denote the inclusion, define $\rho:=\nu\iota:\left[0,1\right]^{2}\rightarrow\left[0,1\right)^{2}$ and denote the quotient topology on $\left[0,1\right)^{2}$ by $\mathcal{O_{\rho}}$.

Then $S:=\left(\left[0,1\right)^{2},\mathcal{O}_{\rho}\right)$ and the topological space mentioned in (a) are homeomorphic.

It is to be shown that $\mathcal{O}_{\nu}=\mathcal{O}_{\rho}$. We have $O\in\mathcal{O}_{\nu}\iff\nu^{-1}\left(O\right)\text{ open}\Rightarrow\rho^{-1}\left(O\right)\text{ open}$ (since $\iota$ is continuous) and next to that $O\in\mathcal{O}_{\rho}\iff\rho^{-1}\left(O\right)\text{ open}$. This allows the conclusion: $\mathcal{O}_{\nu}\subseteq\mathcal{O}_{\rho}$. Note that $\nu^{-1}\left(A\right)=\cup_{n,m\in\mathbb{Z}}\left[\left(n,m\right)+\rho^{-1}\left(A\right)\right]$ for any set $A\subseteq\left[0,1\right)^{2}$ so if $\rho^{-1}\left(O\right)$ is open then so is $\nu^{-1}\left(O\right)$ as a union of open sets. Proved is now that $\mathcal{O}_{\nu}=\mathcal{O}_{\rho}$.

The fact that $\left[0,1\right]^{2}$ is compact implies that its image under the continuous $\rho$ is compact. So $S$ is compact (and so is $T$).

The image under $\nu$ of line $L\subset\mathbb{R}^{2}$ with a slope belonging to $Q$ coincides with the image under $\rho$ of the union of a finite number of closed sets. So this union is closed itself and as a subset of the compact $\left[0,1\right]^{2}$ it is compact. Consequently $\rho$ sends it to a compact set.

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