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.
 A: For a single point $a$, you may be able to manufacture a sequence $d_n$ depending on $v_n$ such that $d_n$ converges to zero. For that point $a$ there would not be ball around a completely contained within the set because you can always find a smaller $d_n$ than the radius of that ball. The set would probably look sort of like a circle with a shrinking radius.
$$
X = \{(x,y) = (r\cos(\theta),r\sin(\theta)), \theta \in (0,2\pi] : x^2 + y^2 < \theta\}
$$
For the point $a=(0,0)$, take any $v=r_v(\cos(\phi),sin(\phi))$ then $d=\frac{r_v \phi}{2}$ so $a + tv < a + dv=dv=\frac{\phi}{2}(\cos(\phi),\sin(\phi))$ and so is in the set.
For any other point this also follows and is the centre of a ball contained in $X$
Suppose $X$ is open, that there is a ball of radius $p$ around $(0,0)$ contained in $X$, well there is some $\theta<p$ such that the point $\frac{p+\theta}{2}(\cos(\theta),\sin(\theta))$ is not in the set $X$ but is in the ball, so the set $X$ is not open.
A: 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. 
