P-adics and Ultrametric Spaces

Prove that the metric space $(\mathbb{R}\left [ t \right ],d)$ is an ultrametric space.

Let $R[t]$ be the set of polynomials in one variable, $t$, with real number coefficients. For any $p\in \mathbb{R}\left [ t \right ]$, let $deg(p)$ be the degree of the polynomial $p$, and define $\left |p \right |=2^{deg(p)}$ and $\left |0 \right |=0$. For any $p, q \in \mathbb{R}\left [ t \right ]$ , define $d(p,q)=\left| p-q \right |$.

Then I believe we can define as follows: A metric space $(R[t],d)$ is an ultrametic space, and $d$ is an ultrametric, if for any $p,q,r$$\in R[t]$, d(p,q) ≤ max⁡ {d(p,r), d(r,q)}.

  • $\begingroup$ Can you explain from where is this problem? The usual definition is $2^{-\deg(p)}$. The ultrametric inequality is as you say, but there are also positivity $d(p,q)\ge 0$, symmetry and $d(p,q)=0$ iff $p=q$. $\endgroup$ – xarles Aug 30 '18 at 13:51
  • $\begingroup$ This definition is without the negative, and you are correct, the other properties need to be shown as well. Thank you. $\endgroup$ – user565684 Aug 30 '18 at 15:21
  • $\begingroup$ You are right, I was confused: the definition is as you wrote. What is exactly your question? $\endgroup$ – xarles Aug 30 '18 at 16:16
  • $\begingroup$ To prove the the metric space is an ultrametric space by proving the Strong Triangle Inequality d(p,q) ≤ max⁡ {d(p,r), d(r,q)} $\endgroup$ – user565684 Aug 30 '18 at 16:37
  • $\begingroup$ Use that $\deg(P + Q) \leq \max(\deg(P),\deg(Q))$. $\endgroup$ – xarles Aug 30 '18 at 16:51

$p-q$ can't possibly have a greater order than both $p$ and $q$, since there is no way to add powers of $t$ to get a power that is greater than all of the terms.

So $deg(p-q)$$max\{deg(p), deg(q)\}$.

$2^x$ is an increasing function, so $a ≤ b$ implies $2^a ≤ 2^b$. Thus $|p-q| ≤ max\{|p|, |q|\}$, as desired.

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