Let $X=\{(x,y)\in\mathbb{R}^2;\;x>0 \;\text{and}\;x^2\leq y\leq2x^2\}$. Prove that for all $(a,v)\in(\mathbb{R^2}\backslash X)\times \mathbb{R}^2$ there exists $\delta>0$ such that $$0\leq t <\delta \Rightarrow a+tv\in \mathbb{R}^2\backslash X$$
This problem is in the section of differentiable functions of my book.
Thanks.
The assertion is true.
Let $a$ in $\mathbb R^2\setminus X$. If $a\ne(0,0)$, $a$ is in $\mathbb R^2\setminus Y$, where $Y=X\cup\{(0,0)\}$ is closed, hence there exists some positive $r$ such that the ball centered at $a$ with radius $r$ does not meet $Y$. Then, for every $v$ in $\mathbb R^2$, $\delta=r/(1+\|v\|)$ proves the claim.
The case when $a=(0,0)$ (which obviously motivates the whole exercise) requires a more specific approach. Note first that $X$ is included in the interior of the positive quadrant $\mathbb R_+^*\times\mathbb R_+^*$. Thus, if $v=(u,w)$ with $u\leqslant0$ or $w\leqslant0$, then every $\delta$ proves the claim since the halfline $\mathbb R_+\cdot(u,w)$ does not meet $X$.
Consider now the case $a=(0,0)$ and $v=(u,w)$ with $u\gt0$ and $w\gt0$. Let $t\gt0$, then $a+tv=(tu,tw)$ and $(tw)/(tu)^2=w/(tu^2)$ hence, if $t\lt w/(2u^2)$ then $(tw)\gt2(tu)^2$, in particular $a+tv$ is not in $X$. Thus $\delta=w/(2u^2)$ proves the claim. (Obviously, in the "interesting" case $a=(0,0)$, no positive $\delta$ works for every unit vector $v$ since every ball centered at $a$ and with positive radius meets $X$.)
