Open set in $\Bbb{R}^2$ Topology Consider $\Bbb{R}^2$ with the usual topology. Let $X$ be a subset of $\Bbb{R}^2$. If for every $a \in X$ and $v \in \Bbb R^2$ there exists a d>0, such that $a+vt\in X$, for every $0\leq t<d$, then X is open.
I suppose this theorem is wrong as the choice of the radius of the open ball $d$ is dependent from choice of the vector $v$. However, I am looking for a counter-example to disprove that.

In the present formulation... 
Let us fix $a=(0,0)$, $X=\{(x,y):x^2+y^2\leq 1\}$. Let $v$ be any unit vector. Your condition is valid for $d=1$, but $X$ is not open. 
