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If $ (X,d) $ is a metric space and $A$ is a subset of $X$, then, for $x$ belonging to $X$, we define $$d(x,A)= \inf \{d(x,a) : a \in A\}.$$ My question is, Why do we use infimum in the definition, and why not minimum, average, supremum, etc.?

user262860
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  • That's a good question! try it yourself: define new $d$ functions with supremum, average and whatever you want,then try to understand what goes "wrong". In other words, you should realize that the functions you define do not satisfy all the properties you would expect from a "distance" function – Giovanni Sep 08 '15 at 00:20
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    The shortest distance from Paris to Russia does not involve the United States... – Carl Mummert Sep 08 '15 at 00:20
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    The desirability of "minimum" should be clear; what's the shortest way to get from $x$ to somewhere in $A$. The fact of the matter is that this minimum is not always attained. So the infimum is the next best thing. – Ian Sep 08 '15 at 00:22
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    The infimum (as a substitute for a minimum that might not exist as explained by Ian above) also has the technical advantage of always being well-defined, whereas averages, suprema and (whatever you else you had in mind by) etc. might not. – Rob Arthan Sep 08 '15 at 00:31

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We can think of $d(x, A)$ as measuring the distance from $x$ to the set $A$: Informally, if we're in Germany and someone asks in English what is the distance to France, probably the person wants to know how far it is to the nearest point on the French border, rather than to the middle (or even further point) of France. (I do not know whether this default sense of distance to a region is uniform across languages.)

Then, one uses the infimum rather than the minimum as the former always exists, whereas the latter does not. Consider, for example, $X = \Bbb R$ endowed with the usual metric $d(x, y) = |x - y|$ and $A = \Bbb R_+ = \{y \in \Bbb R : y > 0\}$. Then, the set $\{d(0, a) : a \in A\}$ is just $A$ itself, but this has no smallest element, and so the minimum of this set does not exist, but its infimum ($0$) does.

Note that one can certainly ask for the average distance between a point $x$ and the points in a set $X$, and sometimes one does want this, but in general one needs more than a metric space structure to do this, namely, something like a measure, which tells us how to weight each part of the metric space $X$ when computing our average.

Travis Willse
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  • @ Giovanni. I think the first condition $d(x,x)=0$ isn't satisfied. – user262860 Sep 08 '15 at 00:30
  • @ Travis. I didn't get your point in the last paragraph about average. Can you explain it mathematically? – user262860 Sep 08 '15 at 00:39
  • @user262860 Are you familiar with the notion of measure space? – Travis Willse Sep 08 '15 at 00:40
  • No. I don't have any idea about this. – user262860 Sep 08 '15 at 00:41
  • If $A$ has just a finite number of points, $a_1, \ldots, a_n$, then one can define the average distance from $x$ to $A$ to be $\bar{d}(x, A) := \frac{1}{n}(a_1 + \ldots + a_n)$. But if $A$ has an infinite number of points, this formula no longer makes sense, and we have to define something like $\bar{d}(x, A) := \frac{\int_A d(x, y) ,d\mu}{\int_A d\mu}$, but this requires a notion of measure $\mu$ on $X$ (or at least on $A$), and different measures lead to different average distances. – Travis Willse Sep 08 '15 at 00:52
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You're standing at $(1,0)$ in $R^2$, looking down on the open interval $(0,\infty)$. Is there a closest point to you? Is there a minimum?

No, any point you pick, there is an even closer one to you than that. ($(x/2,0)$ is closer to you that $(x,0)$ for any $x>0.$)

Closedness of set guarantees existence of a minimizer.

Behnam Esmayli
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