I am told to assume that our dissimilarity measure $d$ satisfies the properties required of it, what seems to be the definition of a metric:
$d(x,y) \geq0 $ and $d(x,y)=0 \Longleftrightarrow x=y$
$d(x,y)=d(y,x)$
$d(x,z)\leq d(x,y) + d(y,z)$
Given that, we define single linkage to be $$d(A,B)=\min_{x\in A, y\in B}\{d(x,y)\}$$ We are told to assume that our dissimilarity measure $d$ satisfies the properties above.
I want to prove (or contradict...) $$d(A,B)=d(B,A)$$
Since I know $d(x,y)=d(y,x)$ it seems like it ought to be straightforward.
Suppose we have $x\in A , y\in B$, then $d(x,y)=d(y,x)$ still holds, as far as I can see.
$\Rightarrow \min\{d(x,y)\}=\min\{d(y,x)\}$
$\Rightarrow \min_{x\in A, y\in B}\{d(x,y)\} = \min_{x\in A, y\in B}\{d(y,x)\} $
$\Rightarrow d(A,B)=d(B,A)$ as required.
Does this constitute an adequate proof? Can someone help me make it a little more rigorous? Is there something obvious I'm missing?