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Let a set of countable points in the closed unit ball in $\mathbb R^n$ be given. Can we find a line $\lbrace tv: v \in \mathbb R^n, t \in \mathbb R \rbrace$ in it that contains an infinite number of these points.

What if one relaxes the condition for a line to allow a smooth curve?

Added I think the idea should be something like this: WLOG assume the points accumulates to the origin, then for each $k$, consider finite points in the annulus $B(0,\frac 1k)-B(0,\frac{1}{k+1})$ and a smooth curve $\gamma_k$ joining them whose minimum speed is a function of $\frac 1k$

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What have you tried? For a line this is easy: note that any line in the plane crosses the unit sphere in the plane (i.e. the unit circle) at most twice. – Gyu Eun Lee Jan 14 '13 at 22:54
@proximal I'm sorry but I don't quite follow... – Apoha Jan 14 '13 at 23:02
Suppose I picked countably many points in the closed unit ball in $\mathbb{R}^2$ specifically so that they all lie in the boundary, i.e. the unit circle. Is there a line in $\mathbb{R}^2$ containing infinitely many of these points? – Gyu Eun Lee Jan 14 '13 at 23:05
Goodness!! Thanks a lot... Clearly, it's time I went to sleep. Btw if you write this out, I'll accept it as an answer – Apoha Jan 14 '13 at 23:32
I believe the answer to the second question is no. I'm imagining a countably infinite set of points on the topologists sine curve with an accumulation point at the origin so that any curve which goes through infinitely many points would have to not be smooth at the accumulation point. (Very interesting problem) – Dan Rust Jan 14 '13 at 23:54

The answer to the first question was hinted at in the comments: If the points are on the unit sphere, a line can only meet at most two of them.

For the relaxation to smooth curves, I think the answer is yes: The set must have an accumulation point since the unit ball is compact. You can assume that it accumulates at $0$, otherwise consider a small ball around some accumulation point and translate it to $0$. Choose a sequence of points $(x_n)_n$ from the set with $\|x_n\| \downarrow 0$ and then choose a subsequence such that $\frac{x_n}{\|x_n\|} \to v \in \mathbb{S}^{n-1}$ ($\mathbb{S}^n$ is compact!). Smoothly connect the $x_n$. Since the $x_n$ lie in cones with apex at $0$ and with opening angle decreasing to $0$, the curve must be differentiable at its endpoint $0$.

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I agree that the set must have an accumulation point. It's not immediately clear that there exists a sequence whose angle tends to some constant though. Would you not need to use the fact that the unit tangent bundle over the disk is also compact? To emphasise why this isn't clear, if you took an infinite set of points on a spiral tending towards 0, you would need to use some clever subset of the points in order for the angle to tend to a constant. That this clever set of points exist isn't obvious. – Dan Rust Jan 15 '13 at 16:28
@DanielR Yes, in fact, the spiral example was one of the reason I asked this question. – Apoha Jan 15 '13 at 19:05
But $\frac{x_n}{\|x_n\|} \in \mathbb{S}^{n-1}$, and $\mathbb{S}^{n-1}$ is compact, so we can choose a convergent subsequence. Or am I missing something? – Thomas Jan 15 '13 at 19:58
Yes @Thomas, I just didn't think you made that very clear. You're really looking for a convergent subsequence in $D^n\times S^{n-1}$ which happens to be compact as it's the product of two compact spaces. You can't consider the points in the disk, and the vectors associated with them seperately as the convergent subsequence each produces may not coincide. Alternatively you can do one, and then the other, in which case the second subsequence you choose will be a subsequence of the first. – Dan Rust Jan 15 '13 at 20:39
@DanielR You are right, I should have written that more clearly. Indeed, I was thinking of first choosing a sequence the norm of which converges monotonically and then choose a subsequence such that the projections to the sphere converge. I will edit my answer. – Thomas Jan 16 '13 at 8:33

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