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Let $(X,d)$ be a metric space and let $A \subseteq X$. We define the distance from a point $x \in X$ to $A$ by $d(x,A)= \inf \{ d(x,a) : a \in A \} $.

What will be the value of $d(x, \emptyset )$? I am confused between $+ \infty $ and $- \infty$. Also, is it possible to find $ \min \{ d(x,a) : a \in \emptyset \} $? (I know $\emptyset$ is empty, just asking symbolically whether we can take $\min$ instead of $\inf$)

Thanks.

leo
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deditus
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    I think the set ${d(x,a):a\in A}$ is itself an empty set. So in that context this question might help. But I could be wrong. – R_D Apr 23 '16 at 06:33
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    Clearly $-\infty$ is impossible since $d:X^2\to [0,\infty)$... – BigbearZzz Apr 23 '16 at 06:34
  • @Rise I don't understand, why will it be an empty set? – deditus Apr 23 '16 at 06:37
  • @BigbearZzz I think $d$ is not necessarily bounded below by $0$ because diameter of an empty set is $ - \infty$. In diameter, we take $sup$ instead of $inf$ – deditus Apr 23 '16 at 06:44
  • @deditus, since you are considering the set of distances from x to no point there are no distances to consider and hence the set is empty. That was my (possibly flawed) logic. – R_D Apr 23 '16 at 06:44
  • @deditus Indeed, some authors agree with the convention that the diameter of an empty set should be $-\infty$, but many hold to $\text{diam}(\emptyset)=0$. I think your question is a matter of convention more than anything else. You should read http://math.stackexchange.com/questions/579373/what-is-the-best-way-to-define-the-diameter-of-the-empty-subset-of-a-metric-spac . – BigbearZzz Apr 23 '16 at 06:56
  • @BigbearZzz The diameter of $\emptyset$ is $0,$ not $-\infty$; it is so defined e.g. in Kuratowski. You are right that $\sup\emptyset=-\infty$ if the supremum is taken in $\mathbb R.$ To get the correct result for $\operatorname{diam}\emptyset,$ you should take the supremum in the complete lattice $[0,\infty].$ This is natural if you think of the codomain of the distance function as being the nonnegative real numbers. – bof Apr 23 '16 at 07:42
  • @bof I also take the stand that $\text{diam}(\emptyset)=0$ but, as I said, there really are some people who use the convention $\text{diam}(\emptyset)=-\infty$, e.g. Simmons and his Introduction to topology. – BigbearZzz Apr 23 '16 at 09:22
  • @BigbearZzz On the other hand, the degree of the zero polynomial really is $-\infty,$ but that's different. – bof Apr 23 '16 at 09:34
  • @bof Agreed. (though I know a certain algebra professor who prefer to left the notion undefined...) – BigbearZzz Apr 23 '16 at 11:07

3 Answers3

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$d(x,\emptyset)=\inf\emptyset=\infty.$ That makes sense because in general, if $A\subseteq B,$ then $d(x,A)\ge d(x,B);$ so $d(x,\emptyset)\ge d(x,B)$ for all $B.$

If the infimum of a set $S$ is attained, that is, if $\inf S\in S,$ then $\min S=\inf S;$ otherwise, $\min S$ does not exist. Thus $\min\emptyset$ does not exist, just as $\min\{\frac1n:n\in\mathbb N\}$ does not exist. It should be no surprise that the empty set has no least element, seeing as it has no elements at all.

bof
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Fix $x \in X$. If $\{d(x,a) \mid a \in \emptyset\} \neq \emptyset$, then there exists $a \in \emptyset$.

It is vacuously true that for each $M \in \mathbb{R}$ we have $a \geq M$, for all $a \in \emptyset$, so it's reasonable to define $inf\{\emptyset\}=\infty$.

The $min$ function returns a minimal element contained in a set for which an order relation is defined, if such an element exists, so it's again reasonable to say $min\{\emptyset\}=\emptyset$.

Wes
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If $A$ is empty, then $\{ d(x,a) : a \in A \}$ is empty. Then inf$\{\emptyset \} = \infty$.

The proof that inf$\{\emptyset \} = \infty$ can be found in this post:

https://math.stackexchange.com/a/432308

Community
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Jack
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