# Prove $(R[t],d)$ is an Ultrametric Space

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)}.

• 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 $p=q$. – xarles Aug 30 '18 at 13:51
• This definition is without the negative, and you are correct, the other properties need to be shown as well. Thank you. – user565684 Aug 30 '18 at 15:21
• You are right, I was confused: the definition is as you wrote. What is exactly your question? – xarles Aug 30 '18 at 16:16
• 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)} – user565684 Aug 30 '18 at 16:37
• Use that $\deg(P + Q) \leq \max(\deg(P),\deg(Q))$. – 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.