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Archive for February 9th, 2009

## 1.3 Exercise 6

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.

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