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1.3 Exercise 6

February 9th, 2009 No comments

I've been a little remiss about writing problems and their solutions out this last week because I'm taking the civil service exam and have been studying for it.  There is lots of history that I don't know!  There is lots of math that I don't know either, but I think I want to aggrandize my knowledge set so that it is diverse, than so specialized.  I favor Encyclopedism. 

Without further ado, however, here's exercise 1.3.6.  This is interesting like the first problem on Equivalence Relations was interesting; ordering based on parabolas.

"Define a relation on the plane by setting

 (x_0, y_0) < (x_1, y_1) if either  y_0 - x_0^2 < y_1 - x_1^2 , or  y_0 - x_0^2 = y_1 - x_1^2 and  x_0 < x_1 .  Show that this is an order relation on the plane, and describe it geometrically."  

(Taken from Topology by James R. Munkres, Second Edition, Prentice Hall, NJ, 2000. Page 29.)

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SOLUTION 

Comparability: Choosing two points  (x_0, y_0) \neq (x_1, y_1) , either  y_0 - x_0^2 < y_1 - x_1^2 or   y_0 - x_0^2 < y_1 - x_1^2 (or both, by the definition of "or": non-reflexivity and transitivity will exclude one or the other possibility).  If they are equal, then  x_0 < x_1 or  x_0 > x_1 (or both).  If  x_0 = x_1 ,  y_0 = y_1 , and then  (x_0, y_0) = (x_1, y_1) , contrary to our assumption.

Non-reflexivity: we want to show that for no  (x, y) in the plane the relation  (x, y) < (x, y) holds.  So suppose it holds for some point.  This means  y^2 - x < y^2 - x , which is not true because of the identity (reflexive) property of equality (an element equals itself).  So then we conclude that  y^2 - x = y^2 - x . At this point we have to check the second criterion,  x < x .  Again this contradicts the identity (reflexive) property of equality.  Our assumption must have been wrong, so there is no  (x, y) that is lesser than itself.

Transitivity.  We want to show that  (x_0, y_0) < (x_1, y_1) and  (x_1, y_1) < (x_2, y_2) \Rightarrow (x_0, y_0) < (x_2, y_2) . There are four cases.  In the first case,  y_0^2 - x_0 < y_1^2 - x_1 and  y_1^2 - x_1 < y_2^2 - x_2 .  This implies  y_0^2 - x_0 < y_1^2 - x_1 < y_2^2 - x_2 , and  y_0^2 - x_0 < y_2^2 - x_2 . This last step is justifiable because the order relation of the reals is axiomatic (hence, transitivity of the reals is axiomatic). In the second case,  y_0^2 - x_0 < y_1^2 - x_1 , and   y_1^2 - x_1 = y_2^2 - x_2 and  x_1 < x_2 .  Substitution shows the implication:  y_0^2 - x_0 < y_2^2 - x_2 .  In the third case,  y_0^2 - x_0 = y_1^2 - x_1 and  x_0 < x_1 , and   y_1^2 - x_1 < y_2^2 - x_2 .  Again substitution shows the implication:  y_0^2 - x_0 < y_2^2 - x_2 .  Finally, in the fourth case,  y_0^2 - x_0 = y_1^2 - x_1 and  x_0 < x_1 , and  y_1^2 - x_1 = y_2^2 - x_2 and  x_1 < x_2 .  This in turn suggests  x_0 < x_1 < x_2 , and  x_0 < x_2 .  Again, this last step is justifiable because we inherit transitivity of the reals from the axiomatic fact that the reals are ordered.

For two points in different standard parabolas  y - x^2 = K , the point that is "lesser" is in the bottomest parabola.  For two points on the same parabola, the one on the left is lesser.  Put a different way, for a single point, such a point is lesser than all the points in parabolas above its own, and lesser than points on the same parabola to the right of itself.