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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?

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    This is not what it means to say that all triangles are isosceles. This means either $d(x, y) = d(x, z)$ or $d(y, x) = d(y, z)$ or $d(z, x) = d(z, y)$. – 2017-01-27
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    As a start, define $|x|$ to be $e^{-\nu_p(x)}$, and prove that $|x+y|$ equals the larger of $|x|$ and $|y|$, unless $|x|=|y|$. – 2017-01-27
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    @GerryMyerson So let $x = p^n u$ and $y = p^m v$ for $n \neq m$ and $u,v \in \mathbb{Z_p}^*$ . $x+y = p^n u + p^m v $. Hence, $| x+y | = e^{-v_p(p^n u + p^m v)}$. From this I can clearly see the property that $v_p(x+y) \geq min\{ v_p(x) + v_p(y) \}$ but the maximum part is still unclear. – 2017-01-27
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    If $a\ge\min\{\,b,c\,\}$, then $e^{-a}\le\max\{\,e^{-b},e^{-c}\,\}$. But more to the point $\nu_p(x+y)=\min\{\,\nu_p(x),\nu_p(y)\,\}$ unless $\nu_p(x)=\nu_p(y)$. – 2017-01-27
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    @GerryMyerson Did you meant to say `max` instead `min` in $v_p(x+y)$ eqn. ? – 2017-01-27
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    $\nu_2(2)=1$, $\nu_2(3)=0$, $\nu_2(2+3)=\nu_2(5)=0=\min\{\,\nu_2(2),\nu_2(3)\,\}$, so I think I meant what I wrote. – 2017-01-27
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    Making any progress, Dark? – 2017-01-28
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    @GerryMyerson I apologize for late response. I have written down what I think should be the proof. If there are any mistakes please point at them. – 2017-01-28

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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.

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    Good. But, what if, say, $\nu_p(x)=\nu_p(y)$? – 2017-01-28
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    Are you still here? – 2017-01-30
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    @GerryMyerson Yes! I am stuck with inequalities but still trying... – 2017-01-31