$\rho\cap{(Y\times{Y})}$ is an equivalence relation on $X\cap{Y}$

Question

I'm having trouble even getting started with this. I have to show that $\rho\cap{(Y\times{Y})}$ is an equivalence relation on $X\cap{Y}$. I suppose elements from $\rho$ are reflexive, symmetric, and transitive. But how to show that the intersection of $\rho$ and $Y\times{Y}$ I don't know....

Answer 1

I'll try to give some intuition as to why its the case and you can formalize it into a proof. You have an equivalence relation, $\rho$, on $X$, so $\rho \subseteq X\times{X}$. Given that $\rho$ is an equivalence relation the elements of the set will satisfy the properties of reflexivity, symmetry, and transitivity. Now introduce another set $Y\times{Y}$ and consider $\rho \cap (Y\times{Y})$. This set will contain those elements in $\rho$ that are also in $Y\times{Y}$. So these elements are in $X\times{X}$ and $Y\times{Y}$ and they satisfy the properties of reflexivity, symmetry, and transitivity. So, $\rho \cap (Y\times{Y})$ is an equivalence relation on $X \cap Y$.

Answer 2

Hint:

• This is not true if $\rho$ can be arbitrary, e.g. consider $\rho = \varnothing$.
• For $\rho$ being an equivalence relation on $X$, instead of $\rho \cap Y^2$ consider $\rho \cap (X \cap Y)^2$ and use $X \cap Y \subseteq X$.

I hope this helps $\ddot\smile$