Let $X$ be the collection of all sequences of positive integers. If $x=(n_j)_{j=1}^\infty$ and $y=(m_j)_{j=1}^\infty$ are two elements of $X$, set

$$k(x,y)=\inf\{j:n_j\neq m_j\}$$ and

$$d(x,y)= \begin{cases} 0 & \text{if $x=y$} \\ \frac{1}{k(x,y)} & \text{if $x \neq y$} \end{cases}$$

I have already shown that $d$ is a metric on $X$. Now, I must prove the following:

(a) $d(x,z) \leq \max\{d(x,y),d(y,z)\}$

(b) Any open ball $D(x,r)$ is a closed subset of $X$.

(c) If $y \in D(x,r)$, then $D(x,r)=D(y,r)$.

(d) If $D(x,r_1) \cap D(y,r_2) \neq \emptyset$, then either $D(x,r_1) \subset D(y,r_2)$ or $D(y,r_2) \subset D(x,r_1)$.

Edit: I have now posted an answer to my own question. Any helpful comments or corrections are welcome.

up vote 5 down vote accepted

(a) Let $x,y,z \in X$. If $x=y$ or $y=z$ or $z=x$, then $d(x,z) \leq \max{ \{d(x,y),d(y,z)\}}$ obviously holds. Otherwise, if $x=(n_j), y=(m_j), z=(k_j)$, then $n_j=m_j$ for $j<k(x,y)$ and $m_j=k_j$ for $j<k(y,z)$. This implies that $n_j=k_j$ for $j<\min\{k(x,y),k(y,z)\}$. Thus, we have that $k(x,z) \geq \min\{k(x,y),k(y,z)\}$ and this is equivalent to writing: $$\frac{1}{k(x,z)} \leq \max \left\{\frac{1}{k(x,y)},\frac{1}{k(y,z)}\right\} \implies d(x,z) \leq \max \left\{d(x,y),d(y,z)\right\}$$ Thus, we have proven that $d$ is an ultrametric space.

(b) Let $y$ be in the boundary of $D(x,r)$. Since $y$ is in the boundary, any open ball $D(y,r)$ must contain points that are in $D(x,r)$. Let $s$ be some number such that $s \leq r$. We now consider the open ball $D(y,s)$. Since $y$ is a boundary point, $D(x,r) \cap D(y,s) \neq \emptyset$, and hence, there must exist some $z \in D(x,r) \cap D(y,s)$. This means that $|z-x|<r$ and $|z-y|<s \leq r$. Applying the result proven in (1), we obtain: $$|y-x| \leq \max\{|y-z|,|z-x|\}<\,\max\{s,r\} = r$$ Then, $y \in D(x,r)$, and we have proven that any boundary point of $D(x,r)$ belongs to $D(x,r)$, which proves that $D(x,r)$ is closed.

(c) By definition, $y \in D(x,r)$ if and only if $|y-x|<r$. Now, let $z$ be such that $|z-x|<r$, and from the result proven in (1), it follows that $|z-y| \leq \max\{|z-x|,|y-x|\}<r$. This implies that $z \in D(y,r)$, which shows that $D(x,r) \subset D(y,r)$. If we were to switch $x$ and $y$ in the above proof, we would obtain that $D(x,r) \supset D(y,r)$, and so we have proven that $D(x,r)=D(y,r)$.

(d) Without loss of generality, assume that $r_1 \leq r_2$. If $D(x,r_1) \cap D(y,r_2) \neq \emptyset$, there exists some $z \in D(x,r_1) \cap D(y,r_2)$. From the result proven in (3), we know that $D(x,r_1)=D(z,r_1)$ and $D(y,r_2)=D(z,r_2)$. Hence, $D(x,r_1)=D(z,r_1) \subset D(z,r_2)=D(y,r_2)$, and so $D(x,r_1) \subset D(y,r_2)$. In the case of $r_1 \geq r_2$, we obtain $D(y,r_2) \subset D(x,r_1)$, as required.

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