I need to show that all triangles in $\mathbb{Z_p}$ are isosceles. Here, $$\mathbb Z_p = \lim_{\longleftarrow} A_n \space, (A_n = \mathbb{Z}/p^n \mathbb{Z})$$
Now, the topology on $\mathbb{Z_p}$ can be defined as $$d(x,y) = e^{-v_p(x-y)}$$ where $v_p$ is the p-adic valuation.
So basically, I think I will be done if I show that for all $x,y,z \in \mathbb{Z_p}$ , we have $$d(x,y) = d(x,z)$$
Now for this I need to show $e^{-v_p(x-y)} = e^{-v_p(x-z)}$ i.e. ${v_p(x-y)} = v_p(x-z)$ .
But I am not able to show this equality. Any help with this! Is there a better way to do this?
As Qiaochu Yuan said in the above comment, we have to show that either $ d(x,y)=d(x,z)$ or $d(y,x)=d(y,z)$ or $d(z,x)=d(z,y)$ holds. Assuming $v_p(x) \neq v_p(y) \neq v_p(z)$.
Assuming $v_p(x) < v_p(y) < v_p(z)$ -
$v_p(x-y) = \min\{v_p(x), v_p(-y)\} = v_p(x) $
$v_p(x-z) = \min(v_p(x), v_p(-z)) = v_p(x)$
Also,
$e^{-v_p(x-y)} = \max\{ e^{-v_p(x)} , e^{-v_p(-y)} \} = e^{-v_p(x)}$
$e^{-v_p(x-z)} = \max\{ e^{-v_p(x)} , e^{-v_p(-z)} \} = e^{-v_p(x)}$
Hence, $d(x,y) = d(x,z)$.
Similarly making other assumptions like $v_p(z) < v_p(x) < v_p(y)$ and $v_p(y) < v_p(z) < v_p(x)$ will give us $d(z,x) = d(z,y) $ and $d(y,z) = d(y,x)$ respectively.