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$\newcommand{\dist}{\mathrm{dist}\,}$ Let $M$ be a metric space and $\emptyset\neq A,B\subset M$ bounded closed subsets. The Hausdorff distance is defined as $$h(A,B)=\max\{\dist(A,B),\dist(B,A)\},$$ where $$\dist(A,B)=\sup_{x\in A}\inf_{y\in B}d(x,y).$$

Why do we define $\dist(A,B)$ in this way? Is't it possible to replace the supremum by the infimum in the definition of $\dist\!$, that is, define $$\dist_{\mathrm{new}}(A,B)=\inf_{x\in A}\inf_{y\in B}d(x,y).$$

What is the impact of this 'new' definition on the 'Hausdorff distance' given by $$h_{\mathrm{new}}(A,B)=\max\{\dist_{\mathrm{new}}(A,B),\dist_{\mathrm{new}}(B,A)\}?$$

  • 3
    One problem that arises if you replace the sup by an inf is that the resulting distance function fails to be a pseudometric, as the triangle property fails to hold. For example, consider the sets $A=\{1\}$, $C = \{-1\}$ and $B=\{z\in\mathbb{C} | |z|=1 \}$ in $\mathbb{C}$ with the usual topology. We then have $d(A,B)=d(B,C)=0$, but $d(A,C)=2$.2012-08-17
  • 3
    Another problem (related to @Old John's observation) is that your suggested distance is zero already if $A$ and $B$ share a point while the Hausdorff distance is a genuine distance function.2012-08-17

2 Answers 2