Distance Between Closed sets in a metric space

[Pbrunett]

Hey guys, thanks for looking at this.

Ok, so we're given the distance, d(x,C) between a point, x, and a closed set C in a metric space to be: inf{d(x,y): for all y in C}. Then we have to generalize this to define the distance between two sets I'm fairly certain you can define it as:

the distance between closed sets D and C in a metric space, d(C,D) = inf{d(y,D): for all y contained in C}. Which should be equivalent to inf{d(x,y): for all x,y contained in C,D respectively}.

My question is this: How to construct an example of two closed, disjoint sets whose distance is zero under this definition? I feel like I need to find two sets containing points that can be made arbitrarily close, but am unsure how to do this without some point being a limit point of both sets, contradicting C,D disjoint.

If you guys have an idea that would be great, I'd much prefer a hint or nudge in the right direction if possible.

Thanks again.

 

[Hurkyl]

I presume you get to pick your metric space too...

 

[Pbrunett]

yes, sorry, forgot to mention that we're talking about a generalized metric space.

I must say that I have a hard time imagining a scenario where this (the aforementioned problem) is possible, I formerly operated under the assumption(read:intuition) that closed and disjoint in a metric space implied some distance between sets.

 

[Hurkyl]

Here's my approach -- figure out your sets first, then decide upon the metric space.

 

[Pbrunett]

thanks for the help!

 

[iluvphysics]

Consider unbounded sets. The distance of two unbounded sets in Euclidean spaces (with the usual metric) can be 0.

Example: Let A = {(t,0): t>=0}, B={(t,1/t): t>=0}. Both are closed, unbounded and their distance is 0.

If one of the sets compact, then the distance can never be zero.

Proof: Let A be compact, B be closed. Since d(A,B) = inf{d(x,y): x in A, y in B}, there exists two sequences (x_n) in A and (y_n) in B, s.t. lim d(x_n,y_n) = d(A,B). Since A is compact, (x_n) has a convergent subsequence (x_n_k), say lim (x_n_k) = x0 in A. Since (d(x_n,y_n)) is convergent, so is its subsequence (d(x_n_k,y_n_k)). In sum,
d(A,B) = lim d(x_n,y_n) = lim (d(x_n_k,y_n_k)) = lim (d(x0,y_n_k)).

Now, if d(A,B) = 0, then x0 is an accumulation point of B. Since B is closed, x0 must be in B. But, x0 cannot be both in A and B: the two sets are disjoint. Thus, we must have d(A,B)>0.