## 1.3 Exercise 1

An easy problem that introduces equivalence relations in somewhat a different scenario than the usual "clock arithmetic" ($\mathbb{Z}$ modulo $n$) context.

"Define two points $(x_0, y_0)$ and $(x_1, y_1)$ of the plane to be equivalent if $y_0 - x_0^2 = y_1 - x_1^2$.  Check that this is an equivalence relation and describe the equivalence classes."

(Taken from Topology by James R. Munkres, Second Edition, Prentice Hall, NJ, 2000. Page 28.)

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SOLUTION

Reflexivity, symmetry, and transitivity of the relation $C$ is but a re-statement of reflexive, symmetric, and transitive properties of equality.

Reflexivity: $\forall x \in A, xCx$ means $\forall (x,y) \in \mathbb{R} \times \mathbb{R}, y + x^2 = y + x^2 \Rightarrow 0 = 0$ is a statement of the identity property of equality.

Symmetry: $xCy \Rightarrow yCx$ means $y_0 - x_0^2 = y_1 - x_1^2 \Rightarrow y_1 - x_1^2 = y_0 - x_0^2$.  This is a statement of symmetry of equality.

Transitivity: $xCy$ and $yCz \Rightarrow xCz$ means $y_0 - x_0^2 = y_1 - x_1^2$  and $y_1 - x_1^2 = y_2 - x_2^2 \Rightarrow y_0 - x_0^2 = y_2 - x_2^2$, which is clear by transitivity of equality.

The family of standard parabolas $y = x^2 + K$ are the equivalence classes ($y - x^2 = K$ describes the parabolic level curves, and hence all points on a particular parabola are in a particular equivalence class).  The statement $y_0 - x_0^2 = y_1 - x_1^2$ is a consequence of transitivity of equality on $K$.

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