In Euclidean space can you always find a sequence approaching a limit point along a line? I have what seems like a very simple question: Suppose I have an open set $X \subset R^n$ and a limit point $z$ of $X$. I would like to find a sequence of points $z_1,z_2,\ldots \subset X$ approaching $z$ along a straight line ; i.e. along the line segment connecting $z_1$ and $z$.
I believe I can do it if $X$ itself is convex but I don't know that. Of course this is all local so if there exists a sequence of open neighborhoods $U_1,U_2,\ldots$ of $z$ such that, for all $k > n_0$, $U_k \bigcap X$ is convex then it should be doable as well but I'm not sure if that's the case (even in Euclidean space topology can be tricky).
Any pointers on how to approach this problem would be appreciated.
Edit: @Apass.Jack has graciously taken the time to shown that in general this is not possible, but is true if $X$ is convex. To see if perhaps I can use that result, I will give an example of an $X$ I had in mind (sorry if this should be a different question, I can create one if necessary). $X$ is the image, by a "projectivization" map $\pi$, of a properly embedded two dimensional surface $L \subset P^3$ (here $P$ is the strictly positive reals so $P^3$ is the strictly positive quadrant of $R^3$). Specifically, if $p \in L$ then $\pi(p) = (p_1,p_2,p_3)/(p_1+p_2+p_3) \in S^2$, where $S^2$ is the two-dimensional simplex. I can show this map, restricted to $L$, is of full rank everywhere, hence is open. I don't know that $X$ is convex, however, which is the problem (seems like it might be, though).
 A: Not necessarily
As you might have suspected, there is not necessarily such a sequence.
Here is a simple counterexample.
Let $X=\{(x,y)\in\Bbb R^2\mid x^2<y<2x^2\}$ and $z=(0,0)$.

Let $\ell$ be a line through $z$.

*

*$\ell$ is line $x=0$ or line $y=0$. Then $\ell$ does not intersect $X$.

*$\ell$ is $y=kx$ for some $k\not=0$. Then the open ball $B(z, |k|/2)\cap X$ does not intersect $\ell$.

However, there is such a sequence if $X$ is convex.
On the other hand, if $X$ is convex, then there is such a sequence.
Here is a brief proof. There are two cases.

*

*If $X$ is on a line, that line must pass through $z$. It is clear that $X$ contain a sequence that approaches $z$.

*Otherwise, there are two points $x_1, x_2\in X$ such that $z, x_1, x_2$ are not colinear. Since $X$ contains $x_1$, $x_2$ and points that can be arbitrarily close to $z$, $X$ contains the interior of the triangle with vertices $z, x_1, x_2$. Hence, $X$ contains the open line segment from $z$ to the midpoint of $x_1$ and $x_2$. There is a sequence of points on that open line segment that approaches $z$.

