Let E = R R be the Euclidean plane. Prove that binary relations S1, S2 are equivalence relations. Also prove that T1, T2 are "very close to be" partial ordering relations on the respective sets, in the following sense: Instead of being antisymmetric, T1 and T2 satisfy the following properties, which resemble antisymmetricity:

({X, Y} i T1 {Y, X} T1) = {X, Y} S1 and ({{x, y}, {x , y}} T2 {{x , y} ,{x, y}} T2) = {{x, y}, {x , y}} S2

Think about it and try to understand why these properties resemble antisymmetricity.

(a) S1 is the binary relation on P(N) defined by {X, Y} S1 X Y is finite

(b) S2 is the binary relation on E defined by {{x, y}, {x , y}} d({x, y},{0, 0}) = d({x , y },{0, 0})

Here d({a, b},{a , b }) is the distance between {a, b} and {a , b}, and recall that d({a, b},{a , b }) = sqrt((a a ) 2 + (b b )^2)

(c) T1 is the binary relation on P(N) defined by hX, Y i T1 X r Y is is finite.

3. (5pt) Let E = R x R be the Euclidean plane. Prove that binary relations S1, $2 are equivalence relations. Also prove that Ti, T2 are "very close to be" partial ordering relations on the respective sets, in the following sense: Istead of being antisymmetric, 71 and 72 satisfy the following properties, which resemble antisymmetricity: ((X, Y) E TI A ( Y, X) ET1) = (X,Y) ES] and (((I, y), (z', y')) ET2 A ((I',y'), (z,y)) ET2) = (((x,y), (1',y'))) ES2 Think about it and try to understand why these properties resemble antisymmetric- ity. (a) S1 is the binary relation on P(N) defined by (X, Y) ES1 XAY is finite (b) S2 is the binary relation on E defined by ((x, y), (x',y')) - d((x,y), (0, 0)) = d((x', y'), (0,0)) Here d((a, b), (a', b'))) is the distance between (a, b) and (a', b'), and recall that d((a, b), (a', b'))) = v(a - a")2 + (6 - b')2. (c) Ti is the binary relation on P(N) defined by (X, Y) ETi X Y is finite. (d) T2 is the binary relation on E defined by ((x, y), (x',y'))