Intersecting Intervals: Consistency of Families F & G

[Diffy]

I define a consistent family, F, to be a set of intervals such that if I_1 and I_2 are in F then I_1 and I_2 intersect

Now given two families F and G, we say F is consistent with G provided that each interval from F intersects with each interval from family G.

My question:
If you know that F is consistent with G, does this imply that either F or G is consistent itself?

My gut feeling is that if F is consistent with G then at either F OR G must be consistent..

 

[HallsofIvy]

Suppose F is a family of vertical intervals from (x, 1) to (x, -1) with x between -1 and 1 and G is a family of horizontal intervals from (1, y) to (-1, y) with y between -1 and 1.

 

[Diffy]

Hey Halls,
Great counterexample!

I am working in a book that is constructing the real number system through these intervals. As such, I have been strictly thinking of these rational intervals as one dimensional along the real number line.

If we restrict ourselves to the real number line then is my assertion true?

 

[Pere Callahan]

If we restrict ourselves to the real number line then is my assertion true?

Wee, I'd say, yes

Proof: edited...proof wrong

 

[Diffy]

There must be something wrong Pere...
If I am reading your proof correctly you are asserting that both F and G are BOTH consistent. However consider F the family {(1,5), (6,10)} and G the family {(3,7), (2, 8)} Here F is consistent with G because each interval in F intersects each interval in G, however F is not consistent.

 

[Pere Callahan]

You're right, my "proof" was dead wrong, gee ... I'll think of sth. else

Next try:

Assume F is consistent with G but neither F nor G is consistent. Clearly, F and G contain more than one interval, otherwise they would be consistent.
For two intervals v,w write v<w if $\forall x\in v, \forall y\in w: x<w$. Our inconsistency assumption allows us to pick $f_1, f_2$ from F and $g_1,g_2$ from G such that

$$f_1\cap f_2 = \emptyset$$
and
$$g_1\cap g_2 = \emptyset$$

w.l.o.g. we take $f_1<f_2$ and $g_1<g_2$. (If neither $f_1<f_2$ nor $f_2<f_1$ were true, there would be $x, x' \in f_1,\quad y,y'\in f_2$ such that x<y and x'>y'; this implies that the number
$$t=\frac{y'(x-y)-y(x'-y')}{(x-y)-(x'-y')}$$
lies between x and x' as well as between y and y'.(I omit the calculation, it is rather obvious that one of two non-intersecting intervals has to be greater than the other one) Hence $t\in f_1 \cap f_2$ which is a contradiction.

F being consistent with G implies


$$f_1\cap g_2 :=h_2 \neq \emptyset$$
$$f_2\cap g_1 :=h_3 \neq \emptyset$$

Choose

$$x\in h_2,\quad y\in h_3$$

There are now two cases:

Case 1: x<=y: Since $x \in g_2$ and $y \in g_1$ this is a contradiction to our assumption $g_1<g_2$.

Case 2: x>y: Since $x \in f_1$ and $y \in f_2$ this is a contradiction to our assumption $f_1<f_2$.

Hence, if F is consistent with G, either F or G or both have to be consistent.

 

[Pere Callahan]

Any objections to this proof?