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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!

Chinnapparaj R
  • 11,589

1 Answers1

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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

$$ra_{22}+sa_{11}+rs+ub_{21}+vb_{12}+uv<\frac{k}{6}+\frac{k}{6}+\frac{k}{6}+\frac{k}{6}+\frac{k}{6}+\frac{k}{6}=k.$$

mfl
  • 29,399