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I Open balls contained in other open balls

  1. Dec 3, 2018 #1
    I am reading a proof that claims that ##\overline{B_{\epsilon /2}(x)} \subseteq B_{\epsilon}(x)##, where the overline means closure, without really explaining why. It's clear to me that ##B_{\epsilon /2}(x) \subseteq B_{\epsilon}(x)##, but I can't quite see why the former is the case without additional explication.
     
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  3. Dec 3, 2018 #2

    fresh_42

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    Can you describe the closure algebraically? Which points are in the closed ball?
     
  4. Dec 3, 2018 #3
    I'm not sure I am able... I might say that the closure of an open ball is a closed ball with same radius, but I know that this is not necessarily true in general metric spaces. So I'm not really seeing how to characterize the closure of the open ball...
     
  5. Dec 3, 2018 #4

    fresh_42

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    The open ball is ##B_\varepsilon(x) =\{\,y\,|\,d(x,y) <\varepsilon\,\}##. The closure includes all limit points, so what do we get if we take the ##y's## to the limits?
     
  6. Dec 3, 2018 #5
    I'm not seeing it... I'm tempted to say that ##\overline{B_\epsilon(x)} = \{y\mid d(x,y)\le \epsilon \}##, but this is just the closed ball, so this can't be right. I don't know what other avenue to take
     
  7. Dec 3, 2018 #6

    fresh_42

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    Yes, that's the closed ball. So where is your problem? ##d(x,y) \leq \dfrac{\varepsilon}{2} < \varepsilon##
     
  8. Dec 3, 2018 #7
    Let's denote the closed ball by ##\overline{B}_\epsilon(x)##.

    So is the idea that ##\overline{B_{\epsilon/2}(x)} \subseteq \overline{B}_{\epsilon/2}(x) \subseteq B_\epsilon(x)##?
     
  9. Dec 3, 2018 #8

    fresh_42

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    I'm not aware of a difference between ##\overline{B_{\varepsilon/2}(x)}## and ##\overline{B}_{\varepsilon/2}(x)##. The idea is simply that everything closer or equal ##\varepsilon/2## is still strictly closer than ##\varepsilon##.
     
  10. Dec 3, 2018 #9
    The difference would be that the closure of an open ball is not necessarily the same as the closed ball with the same radius. This fact is mentioned in the question here https://math.stackexchange.com/ques...sure-of-an-open-ball-equal-to-the-closed-ball
     
  11. Dec 3, 2018 #10

    fresh_42

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    The ##\varepsilon-##notation doesn't make sense in the discrete metric, so I assumed, that you have radii of continuous magnitudes - the usual situation when we consider closed balls. The discrete metric is quite an exception.
     
  12. Dec 3, 2018 #11

    fresh_42

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    For the discrete metric, we have as well the inclusion.

    ## \emptyset \; , \; \{\,x\,\}\; , \; X## are the only sets to be considered here, because as soon as a second point is within a ball, the entire space ##X## is. And ##x## is within the balls, so we only consider ##\{\,x\,\}## and ##X##.

    The only way to get ## B_\varepsilon(x) \subseteq \overline{B_{\varepsilon /2}(x)} ## is to have ## B_\varepsilon(x) =\{\,x\,\} ## and ## \overline{B_{\varepsilon /2}(x)}=X##. The first condition yields ##\varepsilon < 1##, so ##\varepsilon /2 < 1/2## and ## \overline{B_{\varepsilon /2}(x)} = \{\,x\,\} ## and they are equal again.
     
  13. Dec 8, 2018 #12

    WWGD

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    There is a result, possibly for some a definition, in the metric case , that if x is in Class(S) then d(x,S)=0.
     
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