1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
    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


    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
    Well done!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted