prove that {A_a|a?R} is a partion of RxR

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stiffy

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For each a?R, let A_a={(x,y)?RxR|y=a-x^2}. Where R represesnts to the real numbers.

Q: prove that {A_a|a?R} is a partion of RxR

The proof,

i) Let a?R. Then (0,a)?A_a, so A_a??.
ii) Suppose a,b?R and a?b. Now, lets assume (x,y)?A_a?A_b. Then y=a-x^2 and y=b-x^2 by definition. This implies a=b. This is a contradiction. Therefore, A_a?A_b=? when a?b.
iii)Suppose (x,y)?RxR. Choose a=y+x^2?R. Then y=a-x^2?R_a; Therefore, (x,y)?R. Thus, RxR ? ?_(a?R)R_a. Clearly, ?_(a?R)R_a ? RxR.

I can follow the first two parts of the proof. We are just showing that the definition of a partion of some set holds. I just don't understand (iii). Namely, " Choose a=y+x^2?R" and "Thus, RxR ? ?_(a?R)R_a. Clearly, ?_(a?R)R_a ? RxR". I don't see this at all. How do I prove the inclusions? Just pick an arbitrary element in the union and show its an element in the cartesian product? Thats what I would think.

Thanks,
--Dan
 
stiffy said:
For each a?R, let A_a={(x,y)?RxR|y=a-x^2}. Where R represesnts to the real numbers.

Q: prove that {A_a|a?R} is a partion of RxR

The proof,

i) Let a?R. Then (0,a)?A_a, so A_a??.
ii) Suppose a,b?R and a?b. Now, lets assume (x,y)?A_a?A_b. Then y=a-x^2 and y=b-x^2 by definition. This implies a=b. This is a contradiction. Therefore, A_a?A_b=? when a?b.
iii)Suppose (x,y)?RxR. Choose a=y+x^2?R. Then y=a-x^2?R_a; Therefore, (x,y)?R. Thus, RxR ? ?_(a?R)R_a. Clearly, ?_(a?R)R_a ? RxR.

I can follow the first two parts of the proof. We are just showing that the definition of a partion of some set holds. I just don't understand (iii). Namely, " Choose a=y+x^2?R" and "Thus, RxR ? ?_(a?R)R_a. Clearly, ?_(a?R)R_a ? RxR". I don't see this at all. How do I prove the inclusions? Just pick an arbitrary element in the union and show its an element in the cartesian product? Thats what I would think.

Thanks,
--Dan
Click to expand...

For part three, you are showing that any point (x,y) in RxR lies on some parabola A_a.

Letting a = y+x^2, we see that (x,y) is contained in A_a by solving for y: y=a-x^2.

In parts (i) and (ii) you showed that 1) every set A_a is not empty, and 2) They do not overlap. These are the upside-down parabolas centered on the y-axis with verticies (0,a). For each a, we get a new parabola and if a and b are different A_a and A_b do not share a point in common. In part (iii) you are showing that every point (x,y) in RxR lies on one of these parabolas, i.e. (x,y) is a point in A_a for SOME a. Thus RxR is a subset of the union of ALL such parabolas. The second inclusion which is "obvious" is saying that the union of subsets of RxR is a set in RxR.
 
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