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Metric spaces and the distance between sets

  • Thread starter synapsis
  • Start date
  • #1
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Homework Statement


Okay, so we've moved on from talking about R^n to talking about general metric spaces and the differences between the two. We're given that X (a metric space) satisfies the Bolzano-Weierstrass Property and that A and B are disjoint, compact subsets of X. Dist(A,B) is defined as the inf{d(x,y): x in A, y in B}. We're asked to show that Dist(A,B)>0.


Homework Equations


General theory of metric spaces: definition of a metric space, metric, etc.


The Attempt at a Solution


Okay, I think I must be missing something because to me it seems kind of trivial. My proof basically says that, by definition of a metric, d(x,y) is always greater than or equal to 0, with equality holding only when x=y. Since A and B are disjoint in our problem, x does not equal y for all x in A and y in B hence d(x,y) is always greater than 0. Since the inf is the min of these distances, it follows that the inf is always greater than 0.

I didn't use the fact that X satisfies the Bolzano-Weierstrass Property however, which makes me think that I'm missing something. Any help would be greatly appreciated.
 

Answers and Replies

  • #2
362
7
To see where your argument breaks down, consider the open intervals (0,1) and (1,2). Since they are not compact, they don't satisfy the hypothesis of your problem. However, you could use your reasoning to "show" that the distance between them is > 0, which it clearly isn't.

Hint: The sentence "Since the inf is the min of these distances, it follows that the inf is always greater than 0." is the problem with your argument. Try to apply it to the above example.

HTH

Petek
 

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