# This seems impossible (metric space distances)

• Mar 30th 2010, 01:04 PM
Chizum
This seems impossible (metric space distances)
So, in a metric space, we define a distance function as D --> R+ union {0}.

Now, there's a problem in my book where the assumption says

For \$\displaystyle 0 <= x < 1\$, there's a distance (for a sequence) such that \$\displaystyle d(n+1, n+2) < (x) d(n, n+1)\$.

So, for the case where x = 0, we have that \$\displaystyle d(n+1, n+2) < 0\$.

I don't see how this is possible.
• Mar 30th 2010, 01:42 PM
HallsofIvy
Quote:

Originally Posted by Chizum
So, in a metric space, we define a distance function as D --> R+ union {0}.

Now, there's a problem in my book where the assumption says

For \$\displaystyle 0 <= x < 1\$, there's a distance (for a sequence) such that \$\displaystyle d(n+1, n+2) < (x) d(n, n+1)\$.

So, for the case where x = 0, we have that \$\displaystyle d(n+1, n+2) < 0\$.

I don't see how this is possible.

That isn't possible. A "distance" is never negative. Perhaps you have misread it.