1
$\begingroup$

from the actual definition of metric space ,we know that

metric space - a set of points such that for every pair of points there is a nonnegative real number called their distance that is symmetric and satisfies the triangle inequality i am interested what is a symmetric distance?i know triangle equality,something sum of two length is more then third one,but what about symmetric distance?thanks in advance

$\endgroup$

2 Answers 2

6
$\begingroup$

Saying that the metric (or distance) is symmetric just means that the distance from $x$ to $y$ is always the same as the distance from $y$ to $x$. In symbols, for all $x,y\in X$ we have $$d(x,y)=d(y,x)\;.$$

$\endgroup$
8
  • $\begingroup$ thanks in advance,thanks guys $\endgroup$ Commented Sep 18, 2013 at 7:02
  • $\begingroup$ could there be non symmetric distance? $\endgroup$ Commented Sep 18, 2013 at 7:03
  • $\begingroup$ @dato: People have looked at non-symmetric distance functions, because they have real-world applications. In terms of effort, for instance, the ‘distance’ up a hill is longer than the ‘distance’ down that hill if you’re a runner or cyclist. They’re a lot harder to work with, however, and they don’t have nearly so nice an associated body of theory. $\endgroup$ Commented Sep 18, 2013 at 7:05
  • $\begingroup$ but in terms of algebraic does there exist non symmetric distance? $\endgroup$ Commented Sep 18, 2013 at 7:06
  • $\begingroup$ @dato: One can write down asymmetric functions that model the kind of psychological ‘distance’ that I mentioned in the other comment, but they aren’t usually thought of as distance functions; in my (limited) experience they’re more likely to be thought of as measuring energy expenditure, work done, or some psychological analogue of those. $\endgroup$ Commented Sep 18, 2013 at 7:09
1
$\begingroup$

That is, we can't have any $x,y$ in the set such that $d(x,y)>d(y,x)$. We must have $d(x,y)=d(y,x)$ for all $x,y$ in the set. (We want the distance from the one to the other to be the same as the distance from the other to the one, since that's how distance actually works "in real life.")

$\endgroup$

You must log in to answer this question.

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