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Let $X$ be a metric space and let $U,V\subset X$ be open and bounded subsets. Is the Hausdorff metric $$ h(U,V)=\max(\sup_{x\in U}\inf_{y\in V}d(x,y),\sup_{x\in V}\inf_{y\in U}d(x,y)) $$ defined for all possible choices of $U$ and $V$?

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Small typo, want $d(x,y)$ there. –  André Nicolas Sep 2 '12 at 15:21
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Where do you see potential here for something not being defined? –  joriki Sep 2 '12 at 15:21
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The quantity $h(U,V)\in [0,+\infty]$ is defined for any pair of subsets in a metric space. It always satisfies $h(U,U)=0$, $h(U,V)=h(V,U)$, and $h(U,V)\le h(U,W)+h(V,W)$. The issue of being a metric reduces to the following two points:

  1. a metric is required to be finite

  2. a metric is required to be nonzero for distinct elements

You may think that boundedness gives you property 1. But if either of two sets is empty, you get $h(U,V)=\infty$ because the infimum of an empty subset of $\mathbb R$ is $\infty$. Thus, you need both $U$ and $V$ to be nonempty.

Property 2 is more problematic. For example, let $U$ be any nonempty open set and let $V$ be $U$ minus a point. Then $h(U,V)=0$. More generally, you get $h(U,V)=0$ whenever $\overline U = \overline V$. This is because the Hausdorff distance (implicitly) takes into account only the closure of the set. This is why it is normally considered in the context of closed sets rather than open sets.

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