A question about this theorem

[Cratylus]

Theorem
Let A and B be. fully ordered classes. Let (L U)be a partition of A
Prove that(L,U) is a cut of A iff ∀x∈L. and ∀y∈U,x <y.

Since (L,U ) is a cut of A ,I know that L ∩ U=∅
But also (L U) is a partition of A ,so could I write it L ∩ U=∅ ?
or should it be written differently maybe [MATH] L_x \cap U_y = \emptyset ; x\ne y [/MATH] ?

Other than this I can do the proof

 

[HallsofIvy]

L and U are given as sets, but there are NO "\(\displaystyle L_x\)" or "\(\displaystyle U_y\)" mentioned so "\(\displaystyle L_x\cap L_y\)" is meaningless!

 

[Cratylus]

L and U are given as sets, but there are NO "\(\displaystyle L_x\)" or "\(\displaystyle U_y\)" mentioned so "\(\displaystyle L_x\cap L_y\)" is meaningless!

I wrote it that to satisfy the definition of a partition.
So if L and U are given sets and are a partition
[MATH] L \cap U =\emptyset for x\ne y [/MATH]

 

[HallsofIvy]

Yes, but what do "\(\displaystyle L_x\) and \(\displaystyle U_y\) mean? I know that, given some partition of the set of real numbers, there exist a real number, x, such that \(\displaystyle L= L_x= \{p| p< x\}\) and \(\displaystyle U= U_x= \{q| q\ge x\}\) but your use of different x and y is puzzling.

(For a partition of the rational numbers, there does not exist such a rational x. The set of rational numbers is not "complete".)

 

[Cratylus]

Yes, but what do "\(\displaystyle L_x\) and \(\displaystyle U_y\) mean? I know that, given some partition of the set of real numbers, there exist a real number, x, such that \(\displaystyle L= L_x= \{p| p< x\}\) and \(\displaystyle U= U_x= \{q| q\ge x\}\) but your use of different x and y is puzzling.

(For a partition of the rational numbers, there does not exist such a rational x. The set of rational numbers is not "complete".)

Definition Let A be a set; by a partition of A we mean a family {Ai}i∈I of nonempty subsets of A with the following properties:
P1.∀i,j∈I,[MATH] A_i\cap A_j [/MATH]=Ø [MATH] i\ne j [/MATH]
So I substituted x for i and y for j ,I was assuming x and y could be
used to satisfy the definition with the condition that x<y

In doing a rough draft of the proof I didn’t do the latter