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Metric Spaces - Distance Between sets and it's closures

  1. May 2, 2014 #1
    I was trying to prove:
    [itex]d(A,B) = d( \overline{A}, \overline{B} )[/itex]
    I "proved" it using the following lemmas:
    Lemma 1:
    [itex]d(A,B) = \inf \{ d(x,B) \}_{x \in A} = \inf \{ d(A,y) \}_{y \in B}[/itex]
    (By definition we have: [itex]d(A,B) = \inf \{ d(x,y) \}_{x \in A, y \in B}[/itex] )
    Lemma 2:
    [itex]d(x_{0},A) = d(x_{0}, \overline{A})[/itex]

    Proof body:
    [itex]d(A,B) \underbrace{=}_{L1} \inf \{ d(x,B) \}_{x \in A} \underbrace{=}_{L2} \inf \{ d(x, \overline{B}) \}_{x \in A} \underbrace{=}_{L1} \inf \{ d(A,y) \}_{y \in \overline{B}} \underbrace{=}_{L2} \inf \{ d(\overline{A},y) \}_{y \in \overline{B}} \underbrace{=}_{L1} d(\overline{A}, \overline{B})[/itex]

    The problem is that... I have the first lemma proved in the textbook, but the second lemma isn't in it and I couldn't prove it (That's why I said "proved" instead of proved).
    I know that [itex] d( x , \overline{B} ) \leq d(x,B) [/itex] almost trivially because by definition is the infimum of a set and the infimum of a subset must be equal or higher.
    But all the things I though are useful to prove [itex] d( x , \overline{B} ) \leq d(x,B) [/itex] not [itex] d( x , \overline{B} ) \geq d(x,B) [/itex].
    If [itex] d(x,B) = 0[/itex] I know it's true because of a theorem that implies it.
    But if [itex] d(x,B) > 0[/itex] I can't prove [itex] d( x , \overline{B} ) \geq d(x,B) [/itex]
     
    Last edited: May 2, 2014
  2. jcsd
  3. May 2, 2014 #2

    micromass

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    Assume by contradiction that ##d(x,\overline{B})<d(x,B)##.

    Take a sequence ##(x_n)_n## such that each ##x_n\in \overline{B}## and such that ##d(x,x_n)\rightarrow d(x,\overline{B})##. Then there is some ##n## such that

    [tex]d(x,x_n)<d(x,B)[/tex]

    Now try to find a contradiction. What is ##d(x_n,B)##?
     
  4. May 2, 2014 #3
    If [itex]x_{n} \in \overline{B}[/itex] then for every [itex]\varepsilon > 0[/itex], in particular for every [itex]\varepsilon \in (0,d(x,B)-d(x,x_{n}))[/itex] there exists a [itex]y \in B[/itex] such that [itex]d(x_{n},y) < \varepsilon[/itex]
    Then [itex]d(x,y) \leq d(x,x_{n})+d(x_{n},y) < d(x,B)[/itex]
    But [itex]y \in B[/itex], so [itex]d(x,B) \leq d(x,y)[/itex] and we have [itex]d(x,B) < d(x,B)[/itex] which is a contradiction.

    Thank you very much micromass.
     
  5. May 2, 2014 #4

    micromass

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    Well done!
     
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