Question: Two complex numbers are called closed IFF their absolute difference is less than one. Is closeness an equivalence relation.
I'm still fairly uncomfortable dealing with equivalence relation but here's my go at it.
Let $\mathbb{C}=\left \{ a+bi \mid a,b \in \mathbb{Z} \right \}$ Let $R=\left \{ \left ( x,y \right ) \mid \left | x-y \right |<1, x,y \in \mathbb{C} \right \}$
Suppose $a+bi \in \mathbb{C}$. Then, $\left | \left ( a+bi \right )-\left ( a+bi \right ) \right |=0<1 so a+bi \sim a+bi$
Thus, R is reflexivie.
Now, let $\left | \left ( a_{1}+b_{1}i \right )-\left ( a_{2}+b_{2}i \right ) \right | <1$ So, $a_{1}+b_{1}i , a_{2}+b_{2}i$ are both closed.
But, so is $\left | \left ( a_{2}+b_{2}i \right )-\left ( a_{1}+b_{1}i \right ) \right | <1.$ Hence, R is symmetric.
Am I right thus far? How can I demonstrate that R is transitive?
Thanks in advance.
There is no transitivity: $\;1.5\,\leftrightarrow\,1\;$ are closed , and also $\;1\,\leftrightarrow\,0.1\;$ are closed, yet $\;1.5\,\leftrightarrow\,0.1\;$ are not closed.