Proving that $GL(n,\mathbb R)$ is an open subset of $M_n(\mathbb R)$ [the direct way] Definition:  
$GL(n,\mathbb R)$ is the set of all $n \times n$ invertible matrices like $A$ such that $det(A)\neq 0$.  
$M_n(\mathbb R)$ is simply the set of all $n\times n$ matrices. 

Question:

Prove that $GL(n,\mathbb R)$ is an open subset of $M_n(\mathbb R)$ using the definition of an open set  


Note: I know the proof that gets help from the continuity of $det$ function and the fact that $\mathbb R - \{0\}$ is open. But, I want a proof which works just with the direct definition of open subsets (which can be found here).  
Any idea?
 A: The easy way is to observe that $GL(n,\mathbb{R})$ is the preimage of the open set $\mathbb{R}\setminus\{0\}$ under the continuous determinant map.  Since the OP wanted a direct proof, here is one.
In order to show that $GL(n,\mathbb{R})$ is open using the definition of an open set, you need to figure out the radius of an open ball around a matrix which doesn't include a singular matrix.
To be more precise, let's use the $\ell^1$ metric on $GL(n,\mathbb{R})$.  In other words, for $M,N\in GL(n,\mathbb{R})$, 
$$
d_1(M,N)=\sum_{1\leq i,j\leq n}|M_{i,j}-N_{i,j}|.
$$
Fix $M\in GL(n,\mathbb{R})$.  Let $K$ the absolute value of the largest value in $M$.  Let $\epsilon\in M_{\mathbb{R}}(n,n)$ be any matrix such that the absolute value of every entry is at most $1$.  For any $1>\delta> 0$, consider $M+\delta\epsilon$.  Observe that $d_1(M,M+\delta\epsilon)=d_1(0,\delta\epsilon)\leq n^2\delta$.  On the other hand, observe that if $d_1(M,N)<\delta$, then there is some $\epsilon$ so that $N=M+\delta\epsilon$.
For any $N\in B(M,\delta)$, the ball centered at $M$ with radius $\delta$, let $N=M+\delta\epsilon$.  Then, we can calculate the determinant of $N$ as follows ($S_n$ is the permutation group on $n$ elements):
$$
\det(N)=\sum_{\sigma\in S_n}(-1)^{\operatorname{sgn}(\sigma)}\prod_{1\leq i\leq n}N_{i,\sigma(i)}=\sum_{\sigma\in S_n}(-1)^{\operatorname{sgn}(\sigma)}\prod_{1\leq i\leq n}(M_{i,\sigma(i)}+\delta\epsilon_{i,\sigma(i)}).
$$
If you multiply this out, you get $\det(M)$ plus a bunch of terms with a factor of $\delta$.  It's not pleasant to write this out, so I'll use the bounds above.
$$
\det(M)-n!\sum_{i=1}^n\binom{n}{i}\delta^iK^{n-i}\leq\det(N)\leq \det(M)+n!\sum_{i=1}^n\binom{n}{i}\delta^iK^{n-i}.
$$
Since each binomial coefficient is at most $2^n$, we know that (since $\delta<1$) that
$$
\det(M)-n!2^n\max\{1,K^{n-1}\}\delta\leq\det(N)\leq\det(M)+n!2^n\max\{1,K^{n-1}\}\delta.
$$
Therefore, if you choose $\delta<\frac{|\det(M)|}{n!2^n\max\{1,K^{n-1}\}}$, then $\det(N)$ cannot be zero, so the ball of this radius centered at $M$ is entirely within $GL(n,\mathbb{R})$, so $GL(n,\mathbb{R})$ is open.
A: Another approach is to use that
$$
(1 - g)^{-1} = \sum_{n = 0}^\infty g^n
$$
for all $g \in M_n(\mathbb{R})$ with $\lVert g \rVert < 1$ (say we use the operator norm). Substituting $g = gg_0^{-1}$ for $g_0 \in \operatorname{GL}(n, \mathbb{R})$ we find
$$
(gg_0)^{-1} = \sum_{n = 0}^\infty (1 - g g_0^{-1})^n
$$
when $\lVert 1 - gg_0^{-1} \rVert < 1$. Since the operator norm is submultiplicative, this latter condition is implied by $\lVert g - g_0 \rVert < \lVert g_0^{-1} \rVert$. 
We conclude that given $g_0 \in \operatorname{GL}(n, \mathbb{R})$, $g$ is invertible if $\lVert g -  g_0 \rVert < \lVert g_0^{-1} \rVert$ so that the ball around $g_0$ with radius $\lVert g_0^{-1} \rVert$ is included in $\operatorname{GL}(n, \mathbb{R})$. Since $g_0$ is arbitrary, $\operatorname{GL}(n, \mathbb{R})$ is open.
A: Expanding on 3 computational parts of @MichaelBurr 's proof above:
1.
We want bounds for:
$$\det(N)=\sum_{\sigma\in S_n}(-1)^{\operatorname{sgn}(\sigma)}\prod_{1\leq i\leq n}(M_{i,\sigma(i)}+\delta\epsilon_{i,\sigma(i)}).$$

Let $S$ be a non-empty subset of $\{1,2,...,n\}$.$$\prod_{1\leq i\leq n}(a_i+b_i)=\sum_{S\subseteq\{1,2,...,n\}}\left(\prod_{i\in S}a_i\right)\left(\prod_{j\in S^c}b_j\right)$$

For $S\ne\emptyset$,
\begin{align}
\prod_{1\leq i\leq n}(M_{i,\sigma(i)}+\delta\epsilon_{i,\sigma(i)})
&=\sum_{S\subseteq\{1,2,...,n\}}\left(\prod_{i\in S}M_{i,\sigma(i)}\right)\left(\prod_{j\in S^c}\delta\epsilon_{j,\sigma(j)}\right)\\
&=\sum_{S\subset\{1,2,...,n\}}\left(\prod_{i\in S}M_{i,\sigma(i)}\right)\left(\prod_{j\in S^c}\delta\epsilon_{j,\sigma(j)}\right)+\prod_{1\leq i\leq n}M_{i,\sigma(i)}\end{align}
Also since $K,\delta>0$
\begin{align}
\sum_{S\subset\{1,2,...,n\}}\left(\prod_{i\in S}M_{i,\sigma(i)}\right)\left(\prod_{j\in S^c}\delta\epsilon_{j,\sigma(j)}\right)&\le\sum_{S\subset\{1,2,...,n\}}\left(\prod_{i\in S}K\right)\left(\prod_{j\in S^c}\delta\right)\\
&=\sum_{S\subset\{1,2,...,n\}}K^{|S|}\cdot\delta^{(n-|S|)}\\
&=\sum_{i=|S|\\{1\le i\le n-1}}K^i\cdot\delta^{(n-i)}\cdot{\binom{n}{i}}\\
&\le\sum_{i=1}^nK^i\cdot\delta^{(n-i)}\cdot{\binom{n}{i}}\\
\end{align}
where, $|S|$ is the cardinality of $S$. Let $A=\sum_{S\subset\{1,2,...,n\}}\left(\prod_{i\in S}M_{i,\sigma(i)}\right)\left(\prod_{j\in S^c}\delta\epsilon_{j,\sigma(j)}\right)$.
\begin{align}\therefore
\det(N)
&=\sum_{\sigma\in S_n}(-1)^{\operatorname{sgn}(\sigma)}\prod_{1\leq i\leq n}(M_{i,\sigma(i)}+\delta\epsilon_{i,\sigma(i)})\\
&=\sum_{\sigma\in S_n}\left\{(-1)^{\operatorname{sgn}(\sigma)}\sum_{S\subset\{1,2,...,n\}}\left(\prod_{i\in S}M_{i,\sigma(i)}\right)\left(\prod_{j\in S^c}\delta\epsilon_{j,\sigma(j)}\right)\right\}+\sum_{\sigma\in S_n}\left\{(-1)^{\operatorname{sgn}(\sigma)}\prod_{1\leq i\leq n}M_{i,\sigma(i)}\right\}\\
&=\sum_{\sigma\in S_n}\left\{(-1)^{\operatorname{sgn}(\sigma)}\sum_{S\subset\{1,2,...,n\}}\left(\prod_{i\in S}M_{i,\sigma(i)}\right)\left(\prod_{j\in S^c}\delta\epsilon_{j,\sigma(j)}\right)\right\}+\det(M)\\
&=\sum_{\sigma\in S_n}\left\{(-1)^{\operatorname{sgn}(\sigma)}A\right\}+\det(M).
\end{align}
Using
$$-\sum_{\sigma\in S_n}A\le\sum_{\sigma\in S_n}\left\{(-1)^{\operatorname{sgn}(\sigma)}A\right\}\le\sum_{\sigma\in S_n}A$$
and
$$\sum_{\sigma\in S_n}A\le\sum_{\sigma\in S_n}\sum_{i=1}^n K^i\cdot\delta^{(n-i)}\cdot{\binom{n}{i}}\le n!\sum_{i=1}^n K^i\cdot\delta^{(n-i)}\cdot{\binom{n}{i}}=n!\sum_{i=1}^n\binom{n}{i}\delta^iK^{n-i}$$
we finally get,
$$\det(M)-n!\sum_{i=1}^n\binom{n}{i}\delta^iK^{n-i}\leq\det(N)\leq \det(M)+n!\sum_{i=1}^n\binom{n}{i}\delta^iK^{n-i}.$$
2.
We want a simpler form of the bound obtained above, preferably in terms of $\delta$.
$$K^{n-i}\le
\begin{cases}
K^{n-1}(\ge1),  & \text{if $K\ge1$} \\
1(>K^{n-1}) & \text{if $0<K<1$}
\end{cases}\mathrm{~~and~~}\delta^i\le\delta~~(\mathrm{since~}0<\delta<1).$$
Therefore,$$\sum_{i=1}^n\binom{n}{i}\delta^iK^{n-i}\le\sum_{i=1}^n\binom{n}{i}\delta\max\{1,K^{n-1}\}=\delta\max\{1,K^{n-1}\}\sum_{i=1}^n\binom{n}{i}\le\delta\max\{1,K^{n-1}\}2^n$$
Finally,
$$\det(M)-n!\delta\max\{1,K^{n-1}\}2^n\leq\det(N)\leq \det(M)+n!\delta\max\{1,K^{n-1}\}2^n.$$
3.
Now, to show that $N\in GL_n(\mathbb{R})$ it is enough to show that $\det(N)>0$ or $\det(N)<0$. We get this by choosing $\delta$ such that
$$\det(M)-n!\delta\max\{1,K^{n-1}\}2^n>0\mathrm{~~or~~}\det(M)+n!\delta\max\{1,K^{n-1}\}2^n<0$$
$$\Rightarrow\frac{\det(M)}{n!\max\{1,K^{n-1}\}2^n}>\delta\mathrm{~~or~~}\frac{\det(M)}{n!\max\{1,K^{n-1}\}2^n}<-\delta$$
$$\Rightarrow\frac{|\det(M)|}{n!\max\{1,K^{n-1}\}2^n}>\delta.$$
