1
$\begingroup$

Problem:

For points $p = (p_1, p_2)$ and $q = (q_1, q_2)$ in $\mathbb{R}^2$ define:

$d_V(p,q) = \begin{cases}1 & p_1\neq q_1 \ or\ |p_2 - q_2|\geq 1 \\ |p_2 - q_2| & p_1= q_1 \ and\ |p_2 - q_2|< 1 \end{cases}$

Show that $d_V$ is a metric.

My Unfinished Solution:

So there are three states:

$1- \ p_1= q_1 \ and\ |p_2 - q_2|\geq 1 \\ 2- \ p_1= q_1 \ and\ |p_2 - q_2|< 1 \\ 3- \ p_1\neq q_1$

The first and second criterion of a metric space are easy to prove. But I can't gather all the three above-mentioned states for $(p,q,r)$ to prove metricness of $d_V$ for the third criterion, i.e., $d(p, q) + d(q, r) \geq d(p, r)$ for all $p,q,r \in \mathbb{R}^2$.

Thank you.

$\endgroup$
2
$\begingroup$

$\forall p,q : \ d_V(p,q) = |p_2 - q_2| \mathrm \ or \ 1 \ $ and $\ d_V(p,q) \le 1$

Specially,if $ d_V(p,q) =1$ then $d_V(p,q)\le |p_2 - q_2|$

CASE I: $d_V(p,r)=|p_2 - r_2|\ \le |p_2 - q_2|+|q_2 - r_2|$

(Remark: $|p_2 - r_2|\ \le |p_2 - q_2|+|q_2 - r_2|$ is from basic inequation in R: $|a+b|\le |a|+|b|$)

$ \ $

If $|p_2 - q_2|<1$ , $|q_2 - r_2|<1,p_1=q_1,q_1 = r_1$

Then $d_V(p,q)+d_V(q,r)= |p_2 - q_2|+|q_2 - r_2|\ge d_V(p,r)$

$ \ $

Other cases, at least one of equations :$\ d_V(p,q) =1 $ , $d_V(q,r)=1$ holds.

Hence, $d_V(p,q)+d_V(q,r)\ge 1\ge d_V(p,r)$

$ \ $

(Remark: Follow the above argurment, get that $d_V(p,q)+d_V(q,r) = |p_2 - q_2|+|q_2 - r_2| $ or $ \ge 1$)

CASE II: $d_V(p,r)=1 \le |p_2 - r_2|\le |p_2 - q_2|+|q_2 - r_2|$

By the similiar arguement did in CASE I,find $d_V(p,q)+d_V(q,r)\ge d_V(p,r)$

$\endgroup$
3
  • $\begingroup$ Thank you very much for your answer. But still it seems to have problems: Supposing CASE I holds for $(p,r)$, it doesn't mean that each of $(p,q)$ and $(q,r)$ satisfies CASE I. In 4th line you supposed that all $p,q,r$ satisfy CASE II, and in 6th line you supposed $(p,r)$ satisfies CASE I but $(p,q)$ and $(q,r)$ satisfy CASE II. Am I right? $\endgroup$
    – L.G.
    Jan 24 '15 at 8:11
  • $\begingroup$ @Ali.E. I put much more details. $\endgroup$
    – Brian
    Jan 24 '15 at 8:45
  • $\begingroup$ @Syuizen: Would you please guide me how does a open 'ball' look like in this metric? I mean, e.g., for an open 'ball' for taxicab metrican is an open diamond, centered at $p$, with distance $\epsilon $ from $p$ to the corners. Thank you. $\endgroup$
    – user210902
    Feb 1 '15 at 1:01
0
$\begingroup$

Suppose we have three distinct points $x_1 = (p_1,q_1), x_2 = (p_2, q_2), x_3 = (p_3, q_3)$ in the plane (in proving the triangle inequality we can always assume that the three points are distinct; when two or more are equal, it already follows from the other axioms)

We want to show $d_V(x_1, x_3) \le d_V(x_1, x_2) + d_V(x_2, x_3)$.

It's clear from the definition that $d_V(x_1, x_3) \le 1$ always (as is any value of $d_V$). So we can assume (or we are done) that none of the two distances on the right hand side is $1$ and their sum is $<1$ as well. This can only happen if $p_1 = p_2$ and $d_V(x_1,x_2) = |q_1 - q_2| < 1$ and $p_2 = p_3$ and $d_V(x_2, x_3) = |q_2 - q_3| < 1$.

So in particular, $p_1 = p_3$. We know that $|q_1 - q_3| \le |q_1 - q_2| + |q_2 - q_3| = d_V(x_1, x_2) + d_V(x_2, x_3) < 1$ (by assumption), using in the first step the usual inequality for $|\cdot|$, and so $d_V(x_1, x_3) = |q_1 - q_3| \le d_V(x_1, x_2) + d_V(x_2, x_3)$ and we are done.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.