If we instead define a metric as $d:X\times X \rightarrow [0,\infty]$, do we lose any nice properties of metric spaces?

The reason I ask is that I saw this theorem: Given a finite measure space $(X,\mathcal{M},\mu )$, define equivalence relation $A \sim B$ if $\mu(A \mathbin{\Delta} B)=0$ for $A,B\in \mathcal{M}$. Then $d: \mathcal{M} /{\sim} \rightarrow [0,\infty) $ defined by $d(A,B)=\mu(A \mathbin{\Delta} B)$ is a metric.

But the same argument would go through if $X$ wasn't a finite measure space, the only place where it seems to fail is because metrics are defined not to take the value $\infty$.

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    $\begingroup$ Note that \cal changes the font going forward, whereas \mathcal{} applies only to its argument. Other tips: \sim gives $\sim$, things that are binary operations should be spaced correctly via \mathbin, and quotienting by an equivalence relation is denoted with a forward slash. $\endgroup$ May 22 '13 at 23:46
  • $\begingroup$ use $\min\{1,d\}$. $\endgroup$
    – user59671
    May 22 '13 at 23:58
  • $\begingroup$ @ZevChonoles Nice tips! Thank-you! $\endgroup$ May 23 '13 at 0:07

Nothing bad happens if you allow a metric to assume the value $\infty $. In fact, only good things happen.

Nothing bad happens since you can still talk about open balls, the induced topology, uniformly continuous functions, continuous functions, Lipschitz functions etc. and the theory looks pretty much the same as it would if the metric is not allowed to assume the value $\infty $. Moreover, any metric space where infinite distance is assumed naturally breaks down as a disjoint union of metric spaces where infinity is not assumed. The construction is very easy: define an equivalence relation on the points of the space by $x\sim y$ if $d(x,y)<\infty $. The equivalence classes are called galaxies, the space breaks down as the disjoint union of galaxies and each galaxy is metric space where all distances are finite.

That good things happen if you allow infinite distance are related to certain constructions that become nicer. For instance, if you have two metric spaces $M_1,M2$ and you wish to define their disjoint union as a metric space, then the most natural thing to do is require that the distance between a point in $M_1$ and a point in $M_2$ is infinite. If you disallow infinite distances, then you have to sweat unnecessarily to make this construction work. But with infinite distances allowed, the disjoint union of metric spaces becomes the trivial notion it should be. Also, reconciling different metric spaces on the same set becomes more transparent. For instance, suppose that $d_i$ is a family of distance functions on a set $X$. Even if it does not attain infinity, their point-wise supremum might become infinite, and thus fail to be a metric space. But if you allow infinity as a possible distance, then the supremum of any family of distance functions is again a distance function.

The reason that metric spaces with infinity allowed have better closure properties under familiar operations is that the set $[0,\infty ]$ has better closure properties than $[0,\infty )$ does.

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    $\begingroup$ So why is the standard definition of metric without $\infty$? If the theory is the same, it would make sense to be as general as possible in the class of objects it applies to. $\endgroup$ May 23 '13 at 15:51
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    $\begingroup$ Not all definition are optimal at birth. The concept of metric space is rather young (about 100 years old) and we can't ask Frechet anymore. It should be noted that Frechet's work at the time was extremely abstract and was a huge leap forward. Perhaps it was already such a psychological hurdle to consider abstract sets as domains of spaces that the possibility of having infinite distance was not entertained. $\endgroup$ May 23 '13 at 19:37
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    $\begingroup$ I just call it a metric space. I'm not the only one to start accepting $\infty $ into ordinary metric spaces. The book Metric Spaces by Burago et al basically does the same. $\endgroup$ Mar 12 '14 at 3:43
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    $\begingroup$ In other words, the topology your infinity-metric induces is not the same as the one usually assumed for $[0,\infty]$, where $\{\infty\}$ is not open but $(x,\infty]$ is for every $x\in\mathbb R$. $\endgroup$ Feb 10 '17 at 19:06
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    $\begingroup$ I idly came across this question (and your answer) after reading on wiki for metric space itself that the Hausdorff metric can achieve $\infty$ as a value. I came in convinced nothing would be to gain (since composing the metric with a bounded, concave/sigmoid-type function would make it finite-valued with no loss of generality) and there would be no arguments for allowing it, but you convinced me otherwise. $\endgroup$ Jul 16 '20 at 2:47

Instead, think about this. Many measure spaces are $\sigma$-finite; this just means that they are countable unions of subsets of finite measure.

Suppose $(\Omega, \mathcal{S}, \mu)$ is a $\sigma$-finite measure space. Then you can partition it with $\{E_n\}_{n=1}^\infty$ so $\mu(E_n)<\infty$ for all n.

Then for all $n$, $A\mapsto \mu(A\cap E_n)$ is a pseudometric on $\mathcal{S}$. Now put

$$d(A, B) = \sum_{n=1}^\infty {\mu(A \mathbin{\Delta} B)\cap E_n)\over 2^n(1+ \mu(A\ \mathbin{\Delta} B)\cap E_n))}.\qquad A, B \in \mathcal{S}.$$

This is a pseudometric with many of the qualities you are seeking. In fact, it can be seen to be a metric on $\mathcal{M}/\sim$.

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    $\begingroup$ Please, fix the parenthesis in the expression inside the sum. $\endgroup$ Jun 5 '16 at 22:18
  • $\begingroup$ Does this metric have a name? $\endgroup$
    – user683848
    Mar 30 '20 at 7:15

We can use infinite metrics to complete metric spaces in an easy manner:

Given any set $X$, let $\mathbb{R}^X$ be the space of functions $X\to \mathbb{R}$, equipped with the (infinite-valued) uniform metric $d_\infty(f,g)=\sup_{y\in X}|f(y)-g(y)|$. The usual proof shows that $\mathbb{R}^X$ is complete.

Now let $(M,d)$ be a metric space. Set $D\colon M\to \mathbb{R}^M$ as $D(x)=d_x=d(x,\cdot)\colon y\mapsto d(x,y)$ for all $x\in M$. Then $D$ is an isometry: $$|d_x(y)-d_{x'}(y)|=|d(x,y)-d(x',y)|\leq d(x,x')$$ and $$|d_x(x)-d_{x'}(x)|=d(x,x'),$$ which shows that $d_\infty(D(x)-D(x'))=d(x,x')$. The closure $\overline{D(M)}$ is a completion of $M$ (which is unique and satisfies the usual universal property as per general theory).

Note this avoid instances of AC as in the usual construction of a completion (as a quotient of the space of Cauchy sequences on $M$, etc.)


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