TITLE: Is $SL(n,\mathbb{R})/SL(n, \mathbb{Z})$ a Hausdorff space?
QUESTION [6 upvotes]: The special linear group $SL(n, \mathbb{R})$ of degree $n$ over $\mathbb{R}$ is the set of $n \times n$ matrices with determinant $1$, with the group operations of ordinary matrix multiplication and matrix inversion. We denote by $SL(n, \mathbb{Z})$ the group of  $n \times n$ matrices  with integer entries and determinant equals 1. Note that  $SL(n, \mathbb{Z})$  is a discrete subgroup of $SL(n, \mathbb{R})$.
Let $G$ to be a topological group and $H$ to be a subgroup of $G$. We will say that a regular Borel measure $\mu$ on the quotient $G/H$ is a left invariant Haar measure if for all Borel sets $E \subseteq G/H$ and all $g \in G$ we have $\mu(gE) = \mu(E)$. 
If $G$ is a locally compact Hausdorff group and $\Gamma$ is a discrete subgroup such that $G/H$  carries a finite left G-invariant Harr measure, then we say that  $\Gamma$ is a $\textbf{lattice}$ in $G$.
We have the following results:

$SL(n, \mathbb{Z})$ is a lattice in $SL(n,\mathbb{R})$. Moreover, the quotiont  $SL(n,\mathbb{R})/SL(n, \mathbb{Z})$ is not compact. 
(Mahler Criterion) For a sequnece $(g_{m})_{m\in \mathbb{N}}$ of $SL(n,\mathbb{R})$,  the sequence  $(\pi (g_{m}))_{m\in \mathbb{N}}$ of $SL(n,\mathbb{R})/SL(n, \mathbb{Z})$  does not have a convergent subsequence if, and only if, there exists a sequence $v_{m} \in \mathbb{Z}^{n}$ with $v_{m} \neq 0$ such that $g_{m}(v_{m})$ tends to $0$.

where  $\pi :SL(n,\mathbb{R}) \to  SL(n,\mathbb{R})/SL(n, \mathbb{Z})$ is the natural projection. 
The second result suggest to me that $SL(n,\mathbb{R})/SL(n, \mathbb{Z})$  is a Hausdorff space, but I can't find any reference. So, I do not know if it is true. 
Thanks

REPLY [7 votes]: To summarize the comments: Combining Left translation on Lie group of a discrete subgroup is properly discontinuous, Quotient of a topological group by a closed subgroup is Hausdorff, Why is every discrete subgroup of a Hausdorff group closed?
answers your question in affirmative: The quotient is Hausdorff.