Equivalence relations regarding binary relations

Question

Let $R \subseteq X \times X$ be a binary relation for $X = \{a, b, c, d\}$. $R = \{(a, a), (b, c), (c, d), (b, d)\}$. Is the relation an equivalence relation? I don't know if I am proving it correctly by using a formal proof?

Proof:

a)$R \subseteq X \times X$, thus reflexive holds.

b)$R \subseteq X \times X$ then $X \times X \subseteq R$, thus symmetric holds.

c)I don't know how to formally prove it correctly.

Is this correct? Thanks.

Answer 1

No, what you have is not correct.

• For $R$ to be reflexive, it must contain $(x, x)$ for every $x \in X$.
• For $R$ to be symmetric, it must contain $(x, y)$ if it contains $(y, x)$.
• For $R$ to be transitive, it must contain $(x, z)$ if it contains $(x, y)$ and $(y, z)$.

In this case, $R$ is not reflexive or symmetric, but it is transitive.

• Not reflexive: $R$ does not contain $(b, b)$, $(c, c)$ or $(d, d)$.
• Not symmetric: $R$ contains $(b, c)$ but does not contain $(c, b)$. Similarly, $R$ contains $(c, d)$ but not $(d, c)$ and $(b, d)$ but not $(d, b)$.
• Transitive: There is no triple of elements $x, y, z \in X$ such that $R$ contains $(x, y)$ and $(y, z)$ but not $(x, z)$.

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It might help to think of $R$ as defining a binary operator $\sim$: for each pair of elements $x, y \in X$, $x \sim y$ if and only if $(x, y) \in R$.

• The reflexive property means that every element is related to itself: $x \sim x$ for every $x \in X$.
• The symmetric property means that you can switch the order of the operands: $x \sim y \iff y \sim x$.
• The transitive property means that you can "chain" the relation: $x \sim y \text{ and } y \sim z \implies x \sim z$.