Proving Disjointness of a Set Defined by y = a - x^2 for All a in ℝ

[jaqueh]

Homework Statement

For each aεℝ let Aa={(x,y) ε ℝ x ℝ: y = a - x^2}

Homework Equations

Prove that the set {Aa: a ε ℝ}

The Attempt at a Solution

i. let X ε Aa then since X is defined for all aεℝ then X≠∅
ii. let X ε Aa and Y ε Aa, therefore X=(x1,y1) and Y=(x2,y2) and they produce different a values their intersection will be the empty set, thus they are disjoint.
iii. The union over of all elements a in ℝ is all sets x,y in the cross product.

I think my main confusions is i have no idea what X is an element of Aa means. Like is X a pair of x,y or is it a distinct element where the set is defined. I don't know what to do.

 

[Deveno]

each A_a is a curve in the plane, namely: a parabola (opening downwards).

from elementary analytic geometry, it should be clear that each parabola has a vertex at (0,a).

THIS should form the basis of your contention that for any a, A_a is non-empty, since we can, in fact, show that the vertex of the parabola is one of the elements of A_a.

to show that A_a∩A_b = ∅, if a ≠ b, you need to show that:

if (x,y) is in A_a, it is not in A_b, and vice versa.

for part (iii), you need to show that there exists some real number a for ANY pair (x,y), with y = a - x². for example, for (0,0), we can let a = 0.

 

[jaqueh]

ok great i think i get it! i think i was too caught up in trying to make it abide to some formulaic thing and i wasn't thinking outside the box

 

[jaqueh]

And is the equivalence relation for all aεℝ and (x,y) ε ℝxℝ, xRy iff y=a-x²

 

[Deveno]

And is the equivalence relation for all aεℝ and (x,y) ε ℝxℝ, xRy iff y=a-x²

no. we're not relating two real numbers to each other, but pairs of real numbers to OTHER pairs.

so your equivalence relation will have the form:

(x,y) R (x',y') if...?

play with this a little. is (0,0) related to (2,-3)? is (0,1) related to (2,-3)?

i'll give you a hint: you can define R in such a way as to not even mention a.

 

[jaqueh]

no. we're not relating two real numbers to each other, but pairs of real numbers to OTHER pairs.

so your equivalence relation will have the form:

(x,y) R (x',y') if...?

play with this a little. is (0,0) related to (2,-3)? is (0,1) related to (2,-3)?

i'll give you a hint: you can define R in such a way as to not even mention a.

think i got it:
for all (x,y) & (x',y') ε ℝxℝ, (x,y)R(x',y') iff x+y=x'+y'

 

[Deveno]

think i got it:
for all (x,y) & (x',y') ε ℝxℝ, (x,y)R(x',y') iff x+y=x'+y'

well, let's just see if that's true. if it is, then (0,0) should be in the same A_a

(with a = 0) as (-2,2), because:

0+0 = -2+2.

now, 0 = a - 0² means a = 0 (we knew that already, right?)

so 2 = 0 - (-2)² = 4...wait, what?

your formula isn't quite right.

 

[jaqueh]

no. we're not relating two real numbers to each other, but pairs of real numbers to OTHER pairs.

so your equivalence relation will have the form:

(x,y) R (x',y') if...?

play with this a little. is (0,0) related to (2,-3)? is (0,1) related to (2,-3)?

i'll give you a hint: you can define R in such a way as to not even mention a.

well, let's just see if that's true. if it is, then (0,0) should be in the same A_a

(with a = 0) as (-2,2), because:

0+0 = -2+2.

now, 0 = a - 0² means a = 0 (we knew that already, right?)

so 2 = 0 - (-2)² = 4...wait, what?

your formula isn't quite right.

its y-x²=y'-x'²
...
ah i think this is my mind telling me i cannot do math when i am sick

 

[Deveno]

its y-x²=y'-x'²
...
ah i think this is my mind telling me i cannot do math when i am sick

that formula still isn't right. check your signs.

 

[jaqueh]

haha, lol i definitely copied it from my notebook incorrectly
y+x^2=y'+x'^2