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Suppose $A$ is a compact set in $\mathbb{R}^n, n\geq 2$ and it has measure zero with $0\not\in A$, can we find a line $L\subset \mathbb{R}^n$ passing through the origin which intersects $A$ in a finite number of points?

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Why not consider the set $A=\{(x,x)\in\Bbb R^2:\, 1\leq x\leq 2\}$? If a line passing through the origin intersects $A$ at all, it necessarily contains $A$ and therefore intersects $A$ at infinitely many points (unless you don't force the line to be infinite, in which case the question is trivial anyway). –  Clayton Feb 24 '13 at 17:01
    
@Clayton The question is can you always find a line that intersects any such set at finite number of points. Your example does not answer that question. –  Maesumi Feb 24 '13 at 17:17
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@Maesumi: How does it not answer the question? $A$ is compact, has measure zero, doesn't contain the origin, and no line passing through the origin only intersects in a finite number of places. –  Clayton Feb 24 '13 at 17:19
    
As the set is compact it is bounded and so if you were free to choose a line then you could choose it so that it does not cross any point of the set (or its convex hull). The requirement of going through origin is just so that the problem becomes interesting. Other wise it plays no other role. –  Maesumi Feb 24 '13 at 17:23
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so this approach does not work. –  ougao Feb 24 '13 at 17:30

2 Answers 2

up vote 1 down vote accepted

Why not consider the set $\{(x,x)\in\Bbb R^2:\, 1\leq x\leq 2\}$?

This set is clearly compact as it is homeomorphic to $[1,2]$, and any line passing through the origin will either contain $A$ (and therefore intersect $A$ in infinitely many points), or not intersect $A$ at all. Thus, it is not always possible to find a line passing through the origin which intersects a compact set at only finitely many points.

It's easy to see how this generalizes to $\Bbb R^n$, so for any dimension, we have a counterexample to the possibility.

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@daniel things were added later on. –  leo Feb 24 '13 at 17:18
    
I was wrong, I was imaging a square. The comment is now deleted. My objection now is that 0 is a finite number of points. –  leo Feb 24 '13 at 17:25
    
@leo: If that is the only objection, then it is a matter of understanding the question. I took intersecting in $0$ places to not be considered since then it doesn't intersect (matter of phrasing, I suppose). –  Clayton Feb 24 '13 at 17:29
    
Well, one can say that a no intersection is the same as 0 element intersection. It is a question like "is it 0 natural number or isn't? I guess this not the first time this debate occurs :-) –  leo Feb 24 '13 at 17:34

Consider

$$2S^1\cup \bigcup_{k=1}^\infty (2-\frac{1}{k})S^1$$

where $S^1$ is the unit circle for $n=2$. And generally

$$2S^{n-1}\cup \bigcup_{k=1}^\infty (2-\frac{1}{k})S^{n-1}$$

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Thanks, by the way, is the set you defined a compact smooth submanifold in $\mathbb{R}^2$? –  ougao Feb 24 '13 at 20:06
    
@ougao: it is compact, because it is bounded and closed. to show closedness show that any convergent sequence converges in the set. –  user59671 Feb 24 '13 at 20:11
    
I also think this should be a smooth compact submanifold, but it is a homework says that: let $M^n$ be a compact smooth manifold and let $f: M^n\to R^{n+1}$ be a smooth map such that $0\not\in f(M)$. Show that we can find a line $L\subset R^{ n+1}$ passing through the origin which intersects $f(M)$ in a finite number of points. If your example works, then it shows this problem is wrong? –  ougao Feb 24 '13 at 20:15
    
I don't know much about manifolds. However check if connectedness is useful. –  user59671 Feb 24 '13 at 20:22
    
There is no assumption on the connectness in the definition of manifold, maybe I should ask it as another problem. –  ougao Feb 24 '13 at 20:27

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