Finding all metrics of set $X=\{1,2,3\}$ I have the following problem where I'm lost a bit. 

Let $X=\{1,2,3\}$ and $(X,d)$ be a metric space. List all the metrics
  $d$ of $X$ and show that they are equivalent. (Hint: construct a
  $3\times 3$ table)

I'm not sure how to solve this problem, but aren't there infinitely many of the metrics? I mean if $d(x,y) = |x-y|$ then isn't also $d(x,y)=c|x-y|$ allowed metric, where $c\geq0$. The axioms of metric $d$ state that

  
*
  
*$d(x,x)=0$
  
*$d(x,y)=d(y,x)$
  
*$d(x,y)\geq 0$
  
*$d(x,z)\leq d(x,y)+d(y,z)$
  

So all the metrics must fulfill these conditions. If they are equivalent, then they induce the same topology of $\tau$ of $X$, that is the set $\mathcal{B}$ of bases consisting from $r$-balls 
$$B(a,r)=\{\;x\in X\mid d(a,x)<r\}$$
must be equal for all these possible metrics $d$. I'm not sure how I should use the hint to help me. Any ideas? 
Thank you!
 A: Just use the table as they said. This is what you get from the first three
$$\begin{bmatrix}
 & 1 & 2 & 3\\ 
1 &0  & |u| & |v|\\ 
2 & |u| & 0 & |w|\\ 
3 & |v| & |w| & 0
\end{bmatrix}$$
and the last is just a triangle requirement $|u| \leq |v|+|w|$, $|v| \leq |w|+|u|$, $|w| \leq |u|+|v|$ together with the cases when a triangle degenerates into a line (when one of the equation is in the form $|a| = |b|+|c|$) except the trivial cases when any $u=0$ or $v=0$ or $w=0$ because that would imply that two elements are the same.
This covers all possible metrics.


*

*So take a triangle and the distance between vertices. $|u| < |v|+|w|$, $|v| < |w|+|u|$, $|w| < |u|+|v|$
$$
\begin{bmatrix}
 & 1 & 2 & 3\\ 
1 &0  & |u| & |v|\\ 
2 & |u| & 0 & |w|\\ 
3 & |v| & |w| & 0
\end{bmatrix}$$

*Take a line and distance between end points and any point in between.
$$
\begin{bmatrix}
 & 1 & 2 & 3\\ 
1 &0  & |u| & |v|\\ 
2 & |u| & 0 & |u|+|v|\\ 
3 & |v| & |u|+|v| & 0
\end{bmatrix}$$ 
$$
\begin{bmatrix}
 & 1 & 2 & 3\\ 
1 &0  & |u| & |u|+|v|\\ 
2 & |u| & 0 & |v|\\ 
3 & |u|+|v| & |v| & 0
\end{bmatrix}
$$
$$
\begin{bmatrix}
 & 1 & 2 & 3\\ 
1 & 0  & |u|+|v| & |u|\\ 
2 & |u|+|v| & 0 & |v|\\ 
3 & |u| & |v| & 0
\end{bmatrix}
$$
A: I wanted to try also answer this question with my own words. Does it make any sense? Here goes: 

Let $X=\{1,2,3\}$ and $(X,d)$ be a metric space. Because $(X,d)$ is a
  metric space we can find such a set $\mathcal{B}$ of $r$-balls that
  $\mathcal{B}=\{\{1\},\{2\},\{3\}\}$ which are open subsets of $X$. 
Clearly these open subsets of $X$ form a basis for the discrete
  topology of $X$. In other words all the metrics on $X$ are equivalent
  with each other because they all induce the discrete topology on $X$.

