If two balls in this metric space intersect, then one is contained in another where $d(a,b)=\frac{1}{\min(i:n_i\neq m_i)},a\neq b$ Let $\mathbb{N}^{\mathbb{N}}$ be the set of all sequences of positive integers. For $a=(n_1,n_2,\cdots),b=(m_1,m_2,\cdots)$ define $d(a,b)=\frac{1}{\min(i:n_i\neq m_i)}, a\neq b$, $d(a,a)=0$

It can be easily shown that $d(a,b)\leq \max\{d(a,c),d(b,c)\}$, I want to prove that if two balls in this metric space intersect, then one is contained in another
 A: Let $x\in \Bbb{N}^\Bbb{N}$. The ball $B_x(r)$ is equal to all the sequences $z$ such that the first $\lfloor \frac{1}{r}\rfloor$ elements of $z$ coincide with those of $x$.
Let us suppose $B_x(r) \cap B_y(s) \neq \emptyset$, with $r < s$. Then there exists some $z$ such that the first $\lfloor \frac{1}{r}\rfloor$ elements of $z$ coincide with those of $x$ and also the first $\lfloor \frac{1}{s}\rfloor$ elements of $z$ coincide with those of $y$.
Therefore the first $\lfloor \frac{1}{s}\rfloor$ elements of $z$ coincide with those of both $x$ and $y$, so we have $x\in B_y(s)$. Because every element of $B_x(r)$ has its first $\lfloor \frac{1}{r}\rfloor$ elements coinciding with those of $x$, they must also coincide with those of $y$, so $B_x(r)\subset B_y(s)$.
A: HINT: 
Yours is an ultrametric space ( the stronger inequality quoted is satisfied). 
So why don't you prove this fact: if $B = B(x_0, r)$ is a ball in this space and $x$ a point in $B$ then $B = B(x,r)$. Then your statement is clear, the ball of larger radius contains the other. 
