1. ## [SOLVED] Equivalence Relations

Prove that if R is a symmetric, transitive relation on A and the domain of R is A, then R is reflexive on A.

I'm not really sure where to start.

2. Originally Posted by Ryaη
Prove that if R is a symmetric, transitive relation on A and the domain of R is A, then R is reflexive on A.
If $x \in A$ then by the fact that $\mbox{dom}R=A \Rightarrow \quad \left( {\exists y \in A} \right)\left[ {\left( {x,y} \right) \in R} \right]$.
By symmetry we know that ${\left( {y,x} \right) \in R}$.
Now you finish.

3. Originally Posted by Plato
If $x \in A$ then by the fact that $\mbox{dom}R=A \Rightarrow \quad \left( {\exists y \in A} \right)\left[ {\left( {x,y} \right) \in R} \right]$.
By symmetry we know that ${\left( {y,x} \right) \in R}$.
Now you finish.
Would you then go on to say:

Since ${\left( {x,y} \right) \in R}$ and ${\left( {y,x} \right) \in R}$ and since transitivity holds on R, then xRy and yRx $\Rightarrow$ xRx (which is the definition of being reflexive)?

Therefore R is reflexive on A.

4. Originally Posted by Ryaη
Would you then go on to say:

Since ${\left( {x,y} \right) \in R}$ and ${\left( {y,x} \right) \in R}$ and since transitivity holds on R, then xRy and yRx $\Rightarrow$ xRx (which is the definition of being reflexive)?

Therefore R is reflexive on A.
yup

i'd word it to fit the definition more, you know, with quantifiers and all that, but yes, that's the basic idea

5. How could I word this to fit the definition better?

Let $x \in A$. $\mbox{dom}R=A \Rightarrow \left( {\exists y \in A} \right)\left[ {\left( {x,y} \right) \in R} \right]$.
By symmetry we know ${\left( {y,x} \right) \in R}$.
Since ${\left( {x,y} \right) \in R}$ and ${\left( {y,x} \right) \in R}$, then by transitivity we know (xRy and yRx) $\Rightarrow$ xRx. (definition of being reflexive)

Therefore R is reflexive on A.

6. Originally Posted by Ryaη
How could I word this to fit the definition better?

Let $x \in A$. $\mbox{dom}R=A \Rightarrow \left( {\exists y \in A} \right)\left[ {\left( {x,y} \right) \in R} \right]$.
By symmetry we know ${\left( {y,x} \right) \in R}$.
Since ${\left( {x,y} \right) \in R}$ and ${\left( {y,x} \right) \in R}$, then by transitivity we know (xRy and yRx) $\Rightarrow$ xRx. (definition of being reflexive)

Therefore R is reflexive on A.

look at how the definitions are in your text, for instance, they usually end with phrases like "for all $x,y \in R$"

7. It's getting kind of late here. Now that I have the main idea down, I'll work on refining it tomorrow. Thank you for all the help!