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 $M_n(R)$?
The topology of $M_n({\mathbb R})$ is the standard topology of ${\mathbb R}^{n^2}$,
Hint: $\det$ is continuous.