Intuitively I would immediately assume no, but that's not how things usually work in math and considering there are different kinds of infinities I haven't been able to find the answer.

Here's my definition of the distance between 2 sets:

$d(A,B) = \inf{\{||\vec a - \vec b||:\vec a \in A, \vec b \in B\}}$

  • 2
    $\begingroup$ What is a distance for two general sets? Or, do you mean sets in $\Bbb R^n$? $\endgroup$
    – Aahz
    Aug 20, 2014 at 10:33
  • $\begingroup$ This was a question on my exam. The title was all the given information and we simply had to answer yes or no. Though 'our' definition of distance was: $d(A,B) = \inf{\{||\vec a - \vec b||:\vec a \in A, \vec b \in B\}}$ $\endgroup$
    – Joshua
    Aug 20, 2014 at 10:42
  • $\begingroup$ It depends on whether “your” definition of a metric space admits infinite distances. $\endgroup$ Aug 20, 2014 at 12:42
  • 1
    $\begingroup$ Please add the definition of "distance between sets" to the question. I'd do it myself, but you've got the lovely TeX all ready to go in your comment. $\endgroup$ Aug 20, 2014 at 15:38

2 Answers 2


The distance between two sets of the same metric space is defined as:

$$d(A,B) = \inf_{x\in A,\ y\in B} d(x,y)$$

That means that if $x\in A$ and $y\in B$ then $d(x,y) \geq d(A,B)$.

Now, $d(x,y)$ is always finite in a metric space so $d(A,B)$ must be too.


Let $x$ be an element of $A$ and let $y$ be an element of $B$. We know that $\|x-y\|$ is a real number $r$ and so $d(A,B)$ must be at most $r$ by the definition of $d(A,B)$, hence $d(A,B)$ is finite.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .