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Question: Let $L_{n}$ be a line in $\mathbb{R}^2$ for n = 1,2,3... Prove that $\cup_{n=1}^{\infty} L_n \ne \mathbb{R}^2$. (Rudin)

Attempted Answer: My first thought was using the fact that a line segment is a closed set and the union of closed sets is also closed, however that would only work for a finite union of closed sets and I have an infinite union, so I can't use that property. My second thought was trying to use the fact that $\mathbb{R}^2$ is uncountable, but I'm not sure how that would look.

Any suggestions?

Petite Etincelle
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user132039
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  • Consider the set of the slopes of the lines. It is countable. But this is not enough to include the whole plane. – mfl Oct 19 '14 at 20:24
  • Can you expand on why it is not enough to include the whole plane? I'm not sure how your response leads to a solution. – user132039 Oct 19 '14 at 20:26
  • Maybe this is too advanced, but the Baire category theorem says that every complete metric space is a Baire space, that means the countable union of nowhere dense subsets has empty interior. – Stefan Hamcke Oct 19 '14 at 21:44

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Consider the circle $S = \{(x,y) : x^2 + y^2 = 1\}$ in $R^2$, there are uncountable points over the circle.

Each line cuts the circle at most on two points, so the union of countable lines covers at most countable points on the circle

$\cup_{n=1}^{+\infty}L_n$ can't even cover the circle $S$, then how can it cover the whole plane?

I'd like to remark that the conclusion is no longer true if we replace straight lines by arbitrage curves, because of the existence of space-filling curves.

However, the conclusion is still true if the countable curves are of $C^1$ type

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Petite Etincelle
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  • I see. So after concluding that the number of lines in R2 is countable for my problem, I can further conclude that since R2 is uncountable, the lines do not cover all points? – user132039 Oct 19 '14 at 20:28
  • @user132039 the point is that when the number of lines is countable, $S\cap(\cup_{n=1}^\infty L_n)$ is countable – Petite Etincelle Oct 19 '14 at 20:33
  • I understand the concept and I think I understand what you're saying, but I don't know how to prove it formally. – user132039 Oct 19 '14 at 20:39
  • @user132039 which part of the argument is not rigorous for you? – Petite Etincelle Oct 19 '14 at 20:41
  • I don't know how to extend what you're saying from the set S to R2. – user132039 Oct 19 '14 at 20:46
  • @user132039 I've proven there exist points on the circle which don't belong to the union of countable lines. These points are also part of $R^2$, so we've found points of $R^2$ that don't belong to the union of countable lines. – Petite Etincelle Oct 19 '14 at 20:48
  • Oh I see, so using it as a counter example. Thanks that makes more sense. – user132039 Oct 19 '14 at 20:50
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The Lebesgue measure of a line in $\Bbb R^2$ is 0. By countable subadditivity, a countable union of lines must still have measure 0, but the measure of $\Bbb R^2$ is infinite.

doetoe
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