# Thread: This seems impossible (metric space distances)

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

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

I don't see how this is possible.

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

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

I don't see how this is possible.
That isn't possible. A "distance" is never negative. Perhaps you have misread it.