# 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).

• An open set contains an open ball, and an open ball will contain a little line segment. Sep 7, 2022 at 16:30
• Yes, it can live on the boundary, and a ball around that point will have to intersect your open set. Within that intersection you can bubble off a line segment with one end at the limit point. Now trail off points on that line toward the limit point. Sep 7, 2022 at 16:44
• @Randall: Thanks for responding. I tend to agonize over every little detail, though, so let me ask: What exactly do you mean by "bubble off"; IOW, how can you be sure that there exists such a segment? That is, how do you know that, regardless of length or direction, every segment doesn't eventually (as you get closer to $z$) fall completely out of $X$? Put another way, can we be sure there is a line segment issuing from $z$ such that, for every $k$, $U_k \bigcap X$ contains a point of that fixed segment? Sep 7, 2022 at 17:24
• Wow, my intuition is really off. This great answer proves me wrong. Good for you for being suspicious. Sep 7, 2022 at 18:57
• I've been burned too many times, I appreciate all the responders for helping out. Sep 7, 2022 at 19:51

#### Not necessarily

As you might have suspected, there is not necessarily such a sequence.

Here is a simple counterexample.

Let $$X=\{(x,y)\in\Bbb R^2\mid x^2 and $$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 $$z$$.
• Thanks, unfortunately I’ve been trying to prove $X$ is convex to no avail. I suspect it might be, I will edit my original question to describe $X$, perhaps I can do what I want that way. Sep 7, 2022 at 18:45
• Great example. +1 Sep 7, 2022 at 19:01
• Great example. but I feel like this failed because the limit point z wasn't in the set X. If z is in X, then I think it's possible, no? Sep 25, 2022 at 1:55
• @Apass.Jack Could you look at this Dyck path/modular arithmetic problem if you have time? Dec 18, 2022 at 0:37