Projection of a lattice need not be a lattice I have the following problem: Let $L$ be a $\mathbb{R}^n$ lattice (that is a discrete closed $\mathbb{R}^n$ subgroup). Let $E$ be a vector subspace of $\mathbb{R}^n$ and consider $\pi$ to be orthogonal projection on $E$.
Apparently, it is well known that $\pi(L)$ need not be a lattice. I can think of examples of discrete subsets whose projection is not discrete (e.g $\{(n,1/n):n\in\mathbb{N}\}$). But not an example of a lattice.
Does someone know (or can find) an example?
P.S: For completeness, in the case where $L=span_\mathbb{Z}(b_1,\ldots,b_n)$ and $E=(b_i,\ldots,b_n)^\bot$ it is well known that $\pi(L)$ is a lattice since one can "lift" elements from $\pi(L)$ to $L$ and guarantee that the norm doesn't increase much).
Thanks to everyone for your help :)
 A: For subspace $y = x \sqrt 2$  and lattice point $(m,n)$  the projected point is
$$ \left(  \; \; \frac {m+n \sqrt 2}{3} \; , \; \; \; \frac {m \sqrt 2+2n }{3} \; \; \right)  $$
By carefully choosing $m,n$  we can get a point arbitrarily close to the origin
A: This answer expands on my comment and on this question Proving that $m+n\sqrt{2}$ is dense in $\mathbb R$ first.
If $\xi$ is an irrational real, then the set $G$ of all $m + n\xi$, where $m, n \in \Bbb{Z}$ cannot be a lattice in $\Bbb{R}$. (To see this, assume that $G$ is a lattice, so that it has least positive element, say $L = m_1 + n_1\xi$ for some $m_1,n_1 \in \Bbb{Z}$. But then as $\xi$ is irrational, we cam approximate $\xi$ by a rational $a/b$ (with $a, b \in \Bbb{Z}$), such that $0 < a/b - \xi < L$. But that gives us $0 < a - b \xi < L$, contradicting our assumption that $L$ is the least element of $G$.)
Now, returning to the present question. If we take $L = \Bbb{Z}^2$ and $E$ to be the span of $(1, \xi)$ where $\xi$ is any irrational number, then the orthogonal projection of $L$ onto $E$ is the set of all points of the form:
$$\frac{((m, n) \cdot (1, \xi))}{\|(1, \xi)\|^2} (1, \xi) = \frac{(m + n\xi)}{1 + \xi^2}(1, \xi)$$
where $m, n \in \Bbb{Z}$. By the previous paragraph (identifying $\Bbb{R}$ and the span of $(1, \xi)$ in $\Bbb{R}^2$, by the linear mapping that maps  $1$ to $ \frac{(1, \xi)}{\sqrt{1 + \xi^2}}$, the orthogonal projection of $L$ onto $E$ is not a lattice.
