Metric spaces and the distance between sets

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SUMMARY

The discussion centers on proving that the distance between two disjoint, compact subsets A and B of a metric space X, which satisfies the Bolzano-Weierstrass Property, is greater than zero. The definition of distance between sets, Dist(A,B) = inf{d(x,y): x in A, y in B}, is established. A common misconception is that since distances are non-negative and A and B are disjoint, the infimum must also be greater than zero. However, the Bolzano-Weierstrass Property is crucial in ensuring compactness, which guarantees that the infimum is indeed positive.

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  • Understanding of metric spaces and their properties
  • Familiarity with the Bolzano-Weierstrass Property
  • Knowledge of compact sets in topology
  • Basic concepts of distance functions in mathematics
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  • Study the implications of the Bolzano-Weierstrass Property in metric spaces
  • Explore the definitions and properties of compact sets
  • Learn about the construction of distance functions in metric spaces
  • Investigate counterexamples involving non-compact sets and their distances
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Mathematics students, particularly those studying topology and metric spaces, as well as educators looking to clarify concepts related to compactness and distance in metric spaces.

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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.
 
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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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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