Let $f:\mathbb{R^2}\to\mathbb{R^2}$ $\text{be a function given by}$ $f(x,y)=(x^3+3xy^2-15x-12xy,x+y)$. let $S$={$(x,y)\in \mathbb{R^2}:f\; \text{is locally invertible at}(x,y)$}.Then
1).$S=\mathbb{R^2}$ \ $(0,0)$
2).$S$ is open in $ \mathbb{R^2}$
3).$S$ is dense in $\mathbb{R^2}$
4).$\mathbb{R^2}$\ $S$ is countable
solution i tried is
by taking the jacobian of above function
$ J = \left| \begin{align} & 3x^2+3y^2-15 & 6xy-12 \\ & 1 & 1\end{align} \right| = 3x^2+3y^2-15-6xy+12$
now i have to find the points where determinant is zero we put $3x^2+3y^2-15-6xy+12=0$
by solving it we get $(x-y)^2=1$ now we get two lines $x-y=1 \;and\;x-y=-1$
so from this we can see that the points where function is invertible is points not on these lines .
set $S=\mathbb{R^2}$ \ $(x-y)^2=1$
this set is dense so option 3 is true and option 1 is wrong also 4th is wrong.But what about option 2 ?in answer key this right option but i have no idea how this is right .
please help
Thank you.
Option 2) is right because a line in $\mathbb R^2$ is a closed set and therefore the union of two lines is a closed set. Since $S^\complement$ is closed, $S$ is an open set.