Prove: The set $$U:=\Big\{(a,b,c,d) \in \Bbb{R}^4: \vert ad-bc \vert >1\Big\}$$ is open in $\Bbb{R}^4$ .
This question is already asked today [see this post]. The answer in this post involves "inverse image of open set is open under continuity" technique.
My question is: How to prove directly by open set definition ?
Here's my try:
Identify $$\Bbb{R}^4 \sim M_2(\Bbb{R})$$
Then the set can be written as $$U:=\Bigg\{\begin{pmatrix} a &b \\c &d\end{pmatrix}: \vert \text{det} A \vert>1\Bigg\}$$
Our claim is to prove it is open in the standard metric $$d(A,B)=\vert\vert A-B \vert \vert=\sqrt{\sum_{i,j} (a_{ij}-b_{ij})^2}$$ where $A=\begin{pmatrix} a_{11} &a_{12} \\a_{21} &a_{22}\end{pmatrix}$ and $B=\begin{pmatrix} b_{11} &b_{12} \\b_{21} &b_{22}\end{pmatrix}$
Take $A \in U$. Then $\vert \text{det}A \vert>1$. Choose $r=\vert \text{det}A \vert -1>0$. I'm stuck on here.
Is this $r$ works ? If so, how to prove $B(A,r)\subset U$ ?
If not, what's wrong ?
Thanks in advance!
Assume $|b_{ij}-a_{ij}|<\epsilon.$ Then
$$\det(B)-\det(A)=b_{11}b_{22}-b_{12}b_{21}-(a_{11}a_{22}-a_{12}a_{21}).$$
Now
$$b_{11}b_{22}-a_{11}a_{22}=(a_{11}+r)(a_{22}+s)-a_{11}a_{22}=ra_{22}+sa_{11}+rs$$ where $-\epsilon<r,s<\epsilon.$ In a similar way
$$a_{12}a_{21}-b_{12}b_{21}=(b_{12}+u)(b_{21}+v)-b_{12}b_{21}=ub_{21}+vb_{12}+uv$$
where $-\epsilon<u,v<\epsilon.$
Thus
$$\det(B)-(ra_{22}+sa_{11}+rs+ub_{21}+vb_{12}+uv)=\det(A).$$
If $\det(B)=1+k$ we need
$$ra_{22}+sa_{11}+rs+ub_{21}+vb_{12}+uv<k.$$
Consider $\epsilon$ such that $\epsilon^2<\frac{k}{6}$ and $\epsilon\cdot\max\{\max|a_{ij}|,\max|b_{ij}|\}<\frac{k}{6}.$ Thus we have
$$ra_{22}+sa_{11}+rs+ub_{21}+vb_{12}+uv<\frac{k}{6}+\frac{k}{6}+\frac{k}{6}+\frac{k}{6}+\frac{k}{6}+\frac{k}{6}=k.$$