The closure of $\mathbb{Z\times Z\times R}$ in Homeo$(\mathbb R^2)$ is the group of translations of $\mathbb R^2$ This question is related to this question but not the same -
Let $\mathbb{Z\times Z\times R}$ act on $\mathbb R^2$ by $$(m,n,r)\cdot(x,y)=(x+m+r,y+n+r\sqrt{2})$$ I need to prove the following -
If Homeo$(\mathbb R^2)$ is given the compact open topology then the closure of $\mathbb{Z\times Z\times R}$ in Homeo$(\mathbb R^2)$ is the group of all translations of $\mathbb R^2$
My attempt -
Let $f$ be in the closure of $\mathbb{Z\times Z\times R}$. Then there exist a net of points $(f_\alpha)\in\mathbb{Z\times Z\times R}$ (defined by $f_\alpha(x,y)=(x+n_\alpha+r_\alpha,y+m_\alpha+r_\alpha\sqrt{2})$ for some $m_\alpha,n_\alpha\in\mathbb Z$ and $r_\alpha\in\mathbb R$) such that $f_\alpha\longrightarrow f$. So for any $(x,y)\in\mathbb R^2$ we have 
$$f_\alpha(x,y)\longrightarrow f(x,y)$$ $$\Rightarrow(x+n_\alpha+r_\alpha,y+m_\alpha+r_\alpha\sqrt{2})\longrightarrow f(x,y)$$ $$\Rightarrow (n_\alpha+r_\alpha,m_\alpha+r_\alpha\sqrt{2})\longrightarrow f(x,y)-(x,y):=(x_0,y_0)$$
Thus $(n_\alpha+r_\alpha,m_\alpha+r_\alpha\sqrt{2})\longrightarrow (x_0,y_0)$ and by Hausdorffness of $\mathbb R^2$ $(x_0,y_0)$ is independent of the choice of $(x,y)$. So$f$ is given as the translation by $(x_0,y_0)$.
This shows the closure of $\mathbb{Z\times Z\times R}$ is contained in the group of translation of $\mathbb R^2$
Can some one help me with the other way?
If $f$ is given by $f(x,y)=(x+x_0,y+y_0)$ then I need to find a net of points in $\mathbb{Z\times Z\times R}$ converging to $f$. I tried splitting $x_0$ and $y_0$ into integer and fractional part but I'm not sure how to go from there. Suggestions will be appreciated.
Thank you.
 A: Let $T_{v,u}(x,y)=T(x+v,x+u)$ be the translation by $(v,u)$. We want to show that it is in the closure of your family of maps.
Since your group of maps already contains all translation with integer coefficients, we may assume that $v,u \in \mathbb{R}/\mathbb{Z}$. Now choose $r=v+k$ where $k$ is an integer, so that $r\equiv _\mathbb{Z} v$. Thus, modulo $\mathbb{Z}$ your group of actions contains a translation by $(v,v\sqrt{2}+k\sqrt{2})$ for every integer $k$. Check for yourself that the set $\{k\sqrt{2} \mid k\in \mathbb{Z} \}$ is dense in $[0,1]$ modulo $\mathbb{Z}$ so you can come close as you want to $u$. 
Going back to translations in $\mathbb{R}^2$ you get that your group of translations contain the translation by $(v,u+\epsilon)$ where $\epsilon$ can be as small as you want. This sequence of translation will converge (even uniformly) to the translation by $(v,u)$.
In a little bit more geometric way, the action of $\mathbb{R}$ on the torus $\mathbb{R}^2/\mathbb{Z}^2$ sending $(x,y)$ to $(x+r,x+r\sqrt{2})$ has a dense orbit.

EDIT: To show that $\{ \sqrt{2}k \mid k \in \mathbb{Z} \}$, we first prove that for every $n$ there is some $k_n$ such that $\sqrt{2}k_n$ mod $\mathbb{Z}$ is in $(0,\frac {1}{n})$. Once you have such a $k_n$ you get that consecutive points of the form $\sqrt{2}k_n m$ for $m\in\mathbb{Z}$ have distance less than $\frac{1}{n}$. As this is true for every $n$, you can get as close as you want to any point on the unit circle (or equivalently, in $[0,1]$).
Now fix $n$ and consider $k\sqrt{2}$ for $k=0,...,n$. Since $\sqrt{2}$ is not rational, any two such points are distinct modulo $\mathbb{Z}$. Thus, divided the unit circle to n segments of the same length, by the pigeonhole principle, some of these segments contain two points $k_1\sqrt{2},k_2\sqrt{2}$ and your required $k_n$ will be their difference.
