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To abbreviate the expression, "it holds that," I will write "iht."

First a definition. Given a partially ordered set $(P,\geq)$, a closure operator on $P$ is a mapping $\mathrm{cl} : P \rightarrow P$ such that

  1. (Idempotent) For all $d$ iht $\mathrm{cl}(\mathrm{cl}\,d) = \mathrm{cl} \,d.$
  2. (Inflationary) For all $d$ iht $\mathrm{cl} \, d \geq d$
  3. (Non-decreasing) For all $d,d'$ iht if $d \geq d'$, then $\mathrm{cl} \, d \geq \mathrm{cl} \, d'.$

Okay, upwards and onwards. Fix a set $X$, and let $P$ denote the set of all metrics $d$ on $X$. We will allow metrics to take values in $[0,\infty]$. So in particular, infinity is a valid distance. Now partially order $P$ by defining that for all $d,d' \in P$ iht $d \geq d'$ iff for all $x,y \in X$ iht $d(x,y) \geq d'(x,y)$. Finally, let $\mathrm{cl} : P \rightarrow P$ denote the mapping that takes a metric to the induced intrinsic metric. Is $\mathrm{cl}$ a closure operator?

If so, I believe this means that path metric spaces (that is, metric spaces that agree with their induced intrinsic metric) form a complete lattice. The question is then: do any distributive laws hold in this lattice?

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goblin GONE
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  • See http://math.stackexchange.com/q/132404 for a proof of the fact that cl is idempotent provided that $d$ has rectifiable paths connecting any two points. It is then clear that for such metrics cl is a closure operator. –  Apr 02 '13 at 11:43

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The answer by t.b. pointed out by jim assumed the space rectifiably connected, because metric spaces there were understood in the usual way, with finite metric. If finiteness is not required, neither is being rectifiably connected. The operator $\mathrm{cl}$ is indeed idempotent.

I don't understand the lattice part. We have the join operation on path metrics, namely $\operatorname{cl}\max(d,d')$. But what is the meet?

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