# What is the easiest way to show that three lines in two dimensional space do not intersect?

I have two similar questions:

1) Let $X$ and $Y$ be two measure spaces. Suppose for every point $x \in X$ there exists a set $\mathcal{U}_x \subset Y$ of full measure in $Y$. Suppose $V \subset X$ has full measure in $X$. Consider the subset of $X \times Y$ given by $$W := \cup_{x \in V} \{x\} \times \mathcal{U}_x \subset X \times Y$$ Does $W$ have full measure?

2) Same question as above, but $X$ and $Y$ are complex algebraic varieties and we say that a set has full measure if the complement is contained in a finite union of subvarieties of strictly smaller dimension.

The motivation for the question is as follows: How do you rigorously prove this fact: Three generic lines in $\mathbb{C}^2$ do not intersect. The way I want to prove it is using these three facts:

a) Two generic lines in $\mathbb{C}^2$ intersect in only one point.

b) Given a specific point a generic line does not pass through that point.

Can I conclude that three generic lines do not intersect generically simply using a) and b)? The idea being that let $A$ be the space of lines in $\mathbb{C}^2$. Let $$X := A \times A, \qquad Y:= A, \qquad V:= A \times A-\Delta_{A}$$ where $\Delta_A$ is the diagonal of $X$. Now given $(L_1, L_2) \in V$ we define $\mathcal{U}_{L_1,L_2} \in Y$ to be the space of lines that do not pass through the intersection of $L_1$ and $L_2$. The set $W$ as defined earlier is precisely the space of three lines which do not intersect. Why does this set have full measure?

More precisely, what is the easiest way to see this space has full measure, preferably just using the fact that $V$ and $\mathcal{U}_{L_1, L2}$ have full measure.

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Work in the projective space $\mathbb{P}^2$. Then three lines belong to the same pencil if and only if the determinant of the $3 \times 3$ matrix of their coefficients vanishes. This is clearly a closed condition on these coefficients. –  Francesco Polizzi Jul 1 '13 at 15:16
For (1) you apply Tonelli's theorem to the characteristic function of $W$. You need $W$ to be measurable and also either that the measures are $\sigma$ finite or they are complete. –  Bill Johnson Jul 1 '13 at 15:28
Just as an example if $X$ and $Y$ were $\mathbb{R}^n$ with lebeguse measure, is $W$ measurable? –  Ritwik Jul 1 '13 at 15:39
@Franscesco Polizi: This is the question I actually have in mind mathoverflow.net/questions/135436/… –  Ritwik Jul 1 '13 at 16:00
@Bill Johnson: How do we apply Tonelli's theorem to $\chi_{W}$? The set $W$ is not the product of two spaces. –  Ritwik Jul 1 '13 at 16:09

If a line is the set of points of the form $a z + b,$ where $a, b$ are complex vectors and $z$ is the complex parameter, the condition that three such lines intersect will be a condition that a certain polynomial in the parameters ($a_i, b_i$ for $i=1, 2, 3$) vanishes (it will be a determinant), so by your definition, the complement of the vanishing locus will be generic. Bringing in measure theory is probably a mistake.

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The answer to question $2$ is yes, if we know the following continuity condition: that the complement of $W$ is itself a union of varieties. Then the complement is just the union of $V^\perp \times Y$ with another variety whose codimension is positive since it's codimension is positive in each fiber, using the formula $dim(totalspace) \leq dim(base)+ dim(largest fiber)$

It is no without this continuity condition. Take $X = \mathbb C$, $Y = \mathbb C$, $\mathcal U_x = \mathbb C - \{ e^x\}$, $V = \mathbb C$.

Since the continuity condition obviously holds in these sorts of (sufficiently simple) algebraic geometry moduli problems, you can deduce the desired result.

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Can you give me a reference for this fact: "dim(totalspace)≤dim(base)+dim(largestfiber)"? Secondly, by union you mean finite union? –  Ritwik Jul 1 '13 at 18:06
Is it immediately obvious that the $W$ in my example of three lines satisfies that continuity condition? –  Ritwik Jul 1 '13 at 18:17
1. For what notion of dimension? 2. yes, because the first-order theory of algebraically closed fields has quantifier elimination. –  Will Sawin Jul 2 '13 at 14:15

For question 1.

As Bill Johnson noted, if $W$ is measurable, then the answer is "yes". But (unless some hypothesis is added on how $\mathcal U_x$ depends on $x$) it need not be the case that $W$ is measurable. We may conclude, anyway, that $W$ has full outer measure.

Sierpinski (I believe) showed (assuming the Continuum Hypothesis) that there is a set $W \subseteq \mathbb R \times \mathbb R$ such that all vertical cross-sections $$\mathcal U_x = \{ y: (x,y) \in W\}$$ have countable complement, while all horizontal cross-sections $$\mathcal U^y = \{ x: (x,y) \in W\}$$ are countable. In this case, $W$ is a badly non-measurable set.

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For instance such a $W$ may be the (graph of) any order relation on $\mathbb{R}$ that is isomorphic to the well-order of $\omega_1$ –  Pietro Majer Oct 26 '13 at 13:18