Is $GL(n;R)$ closed as a subset of $M_n(R)$?

Question

Let $M_n(R)$ denote the space of all $n×n$ matrices with real entries. The general linear group over real numbers,denoted $GL(n,R)$, is given by $GL(n,R)=${$A∈M_n(R)|det(A)\neq0$}. Is $GL(n,R)$ closed as a subset of $M_n(R)$?

Some thought of mine:Obviously,there exists a sequence $A_n$ in $GL(n,R)$ such that $A_n$ tends to $A$ as $n$ to infinity where $A$ is not invertible.For instance,considering $n=2$,let $A_n=\begin{pmatrix} 1 && 0\\ 0 && \frac{1}{n}\end{pmatrix}$.Then $limA_n=A=\begin{pmatrix} 1 && 0\\ 0 && 0\end{pmatrix} ,n\to{+\infty}$,where $det(A)=0$.The example tells us that $GL(n,R)$ is not closed in $M_n(R)$.

How can we prove $GL(n,R)$ is open in $M_n(R)$ directly and strictly?And how do you define the topology in $M_n(R)$?

$M_n(\mathbb{R})$ is - topologically - just a fancy way of saying $\mathbb{R}^{n^2}$. The topology is the usual topology induced by the Euclidean metric (or any other norm, they are all equivalent). Then you can say that the determinant is continuous (it's a polynomial), hence $GL_n(\mathbb{R}) = \det^{-1} (\mathbb{R}\setminus{0})$ is open. — Daniel Fischer, Oct 22 '13 at 15:09

score 7 · Accepted Answer · answered Oct 22 '13 at 15:09

7

The topology of $M_n({\mathbb R})$ is the standard topology of ${\mathbb R}^{n^2}$,

Hint: $\det$ is continuous.

answered Oct 22 '13 at 15:09

Robert Israel

448,999

Can you prove this directly? (without using the fact that det is continuous and just by definition) – Arman Malekzadeh Apr 28 '17 at 13:12

score 1 · Answer 2 · edited Jun 26 '19 at 06:26

Put another way, if you give $M_{n}(\mathbb{R})$ the Euclidean topology, then one may identify it with $\mathbb{R}^{n^{2}}$, which is - with the Euclidean topology - a connected space. If the general linear group were closed, then since it also open (by the continuity of the determinant function), we obtain a contradiction (connected spaces cannot contain any proper subsets which are both open and closed).

But, really, it's much faster to do as the previous poster said, and just conclude that the complement of the general linear group is closed by the continuity of the determinant function and the fact that point sets are closed in the (Hausdorff) Euclidean topology on $\mathbb{R}$.

Is $GL(n;R)$ closed as a subset of $M_n(R)$?

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