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Problem 10.2. Let R be a relation on the set X (so $R \subseteq X \times X$). a) Show that the following are equivalent: (i) The relation R is both symmetric and anti-symmetric. (ii) We have $R \subseteq \Delta_X := \{(x, x) \mid x \in X\}$. b) Show that the following are equivalent: (i) The relation R is both an equivalence relation and a partial ordering. (ii) We have $R = \Delta_X$. c) Show: if X has at least two elements, there is no relation on R that is both an equivalence relation and a total ordering.

          Problem 10.2. Let R be a relation on the set X (so $R \subseteq X \times X$).
a) Show that the following are equivalent:
(i) The relation R is both symmetric and anti-symmetric.
(ii) We have $R \subseteq \Delta_X := \{(x, x) \mid x \in X\}$.
b) Show that the following are equivalent:
(i) The relation R is both an equivalence relation and a partial ordering.
(ii) We have $R = \Delta_X$.
c) Show: if X has at least two elements, there is no relation on R that is both
an equivalence relation and a total ordering.
        
Show more…
Problem 10.2. Let R be a relation on the set X (so R ⊆ X × X).
a) Show that the following are equivalent:
(i) The relation R is both symmetric and anti-symmetric.
(ii) We have R ⊆ := {(x, x) | x ∈ X}.
b) Show that the following are equivalent:
(i) The relation R is both an equivalence relation and a partial ordering.
(ii) We have R =.
c) Show: if X has at least two elements, there is no relation on R that is both
an equivalence relation and a total ordering.

Added by Rita W.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Problem 10.2. Let R be a relation on the set X (so R ⊆ X × X) a) Show that the following are equivalent: i) The relation R is both symmetric and antisymmetric. ii) We have R_x = {x | x ∈ X} b) Show that the following are equivalent: (i) The relation R is both an equivalence relation and a partial ordering (ii) We have R = x. c) Show: if X has at least two elements, there is no relation on R that is both an equivalence relation and a total ordering.
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Transcript

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00:02 Right so solution array for x y in r2 then x is one time x y is one time y so choose k is equal to one we have x y star x y so star is reflexive but me it is symmetric so for one two okay so x1 x2 to be 2 4 and 1 2 we have 1 is 1 half time 2 2 is 1 half time 4 so 2 4 star 1 2 however 1 2 star 2 4 does not hold because 2 is 2 times 1 4 is 2 times 2 therefore star is not symmetric but c is the anti -symmetric if x1 x2 star y1 y2 then y1 is kx1 y2 is kx2 and suppose that y1 x1 x2 is different from y1 y2 now the condition is 0 less than k less than 1 equal to 1 therefore x1 is 1 over k y1 x2 is…
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