Consider $A=\{a,b,c,d,e\}$ in $\mathcal P(A)$ .
And let $XR_1Y$ if and only if there's a bijection of $X$ on $Y$.
How can i proof/know if is transitive, symmetric and reflexive?
Consider $A=\{a,b,c,d,e\}$ in $\mathcal P(A)$ .
And let $XR_1Y$ if and only if there's a bijection of $X$ on $Y$.
How can i proof/know if is transitive, symmetric and reflexive?
This relation is reflexive, symmetric, and transitive:
For any $X$, the following function is a bijection from $X$ to $X$: $f(x) = x$ for all $x \in X$. So: $XRX$ for any $X$. So, $R$ is reflexive.
For any $X$ and $Y$: if $XRY$ then there is a bijection $f$ from $X$ to $Y$, but that means that $f^{-1}$ is a bijection from $Y$ to $X$. So, $YRX$. So, $R$ is symmetric.
For any $X$, $Y$, and $Z$: if $XRY$ and $YRZ$, then there is a bijection $f$ from $X$ to $Y$ and a bijection $g$ from $Y$ to $Z$. But that means that the composition function $fg$ is a bijection from $X$ to $Z$. So, $XRZ$. So, $R$ is transitive.