6
$\begingroup$

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:

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

$\endgroup$
  • 2
    $\begingroup$ What is a distance for two general sets? Or, do you mean sets in $\Bbb R^n$? $\endgroup$ – Aahz Aug 20 '14 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 '14 at 10:42
  • $\begingroup$ It depends on whether “your” definition of a metric space admits infinite distances. $\endgroup$ – Incnis Mrsi Aug 20 '14 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. $\endgroup$ – Kyle Strand Aug 20 '14 at 15:38
12
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.