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&lt;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&lt;f_2 and g_1&lt;g_2. (If neither f_1&lt;f_2 nor f_2&lt;f_1 were true, there would be x, x&#039; \in f_1,\quad y,y&#039;\in f_2 such that x<y and x'>y'; this implies that the number
<br /> t=\frac{y&#039;(x-y)-y(x&#039;-y&#039;)}{(x-y)-(x&#039;-y&#039;)}<br />
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&lt;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&lt;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?