Closed set, compact set, and a definition of distance between sets

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SUMMARY

The discussion focuses on the mathematical concepts of closed sets and compact sets within the context of distance between sets in R^n. It establishes that for two closed sets E and F, the distance defined as d(E,F) can equal zero while their intersection remains a null set. However, when both E and F are compact sets and d(E,F) equals zero, it is proven that their intersection cannot be null. This conclusion is supported by the properties of compactness and convergence of sequences within these sets.

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  • Understanding of closed sets in topology
  • Familiarity with compact sets in R^n
  • Knowledge of distance metrics in Euclidean spaces
  • Basic principles of convergence of sequences
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  • Learn about compactness in metric spaces
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Homework Statement



Let E and F be 2 non-empty subsets of R^{n}. Define the distance between E and F as follows:

d(E,F) = inf_{x\in E , y\in F} | x - y |

(a). Give an example of 2 closed sets E and F (which are non-empty subsets of R^n) that satisfy d(E,F) = 0 but the intersection of E and F is a null set.

(b). If E and F (non-empty subsets of R^n) are compact sets and d(E,F) = 0, prove that the intersection of E and F cannot be the null set.


Homework Equations



The Attempt at a Solution



Part (b) of the question suggests that the closed sets in part (a) are not bounded. But I still can't find these 2 sets with d(E,F) = 0 and yet they don't intersect.

For part (b), maybe I can find a sequence of points x_{k} in E and another sequence y_{k} in F such that as k increases, the distance between x_{k} and y_{k} decreases. Then since E and F are compact, therefore these 2 sequences must converge. And since d(E,F) = 0, hence these 2 sequences must converge to the same point. And then since E and F are compact, this limit point must lie in both E and F, thus E and F share at least one common point.
 
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For part (b), think about the graphs of 1/x2 and -1/x2 as an example.

That looks like a pretty good sketch argument. Only thing is that only a subsequence of each of those must converge, but it amounts to the same result here
 

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