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Need help with simple proof, metric space, open covering.

  1. Jan 13, 2014 #1
    Please take a look at the proof I added, there are some things I do not understand with this proof.

    1. Does it really show that |f(x)-f(y)|≤d(x,y) for all x and y? Or does it only show that if there is an ball with radius r around x, and this ball is contained in an O in the open covering, and if y is contained in that ball. Then |f(x)-f(y)|≤d(x,y)?

    2. Also please take a look at the sentence I underlined with a red line. Why can he say that
    f(y)≥f(x)-d(x,y)? I could understand if he wrote r instead of f(x), that is f(y)≥r-d(x,y), why is he allowed to use r instead of f(x)?
     

    Attached Files:

  2. jcsd
  3. Jan 13, 2014 #2

    jbunniii

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    In the second paragraph, there is an implicit assumption that ##d(x,y) < r##. This assumption is fine since continuity at ##x## depends only on what happens near ##x##, but this should have been stated explicitly, because as written, the rest is nonsense if ##d(x,y) \geq r##.

    Regarding the red underlined inequality: I agree with you that the immediate implication is ##f(y) \geq r - d(x,y)##. This inequality is valid for any ##r## in the set ##S = \{r \in \mathbb{R} | r < 1 \text{ and } d(x,y) < r \text{ and } B(x;r) \subseteq O\}##. Therefore it is also valid for ##r = \sup S##, but ##\sup S## is simply ##f(x)##, by definition.
     
  4. Jan 13, 2014 #3
    Thank you very much! But what is it that makes the inequality still hold? What has me puzzled is that r ≤ sup S for any R in S? So it seems to me that we can't change sup S with r, since the inequality in
    f(y) ≥ r - d(x,y) goes the other way?

    I mean, if we have a change a part of the smaller size of the inequality with a bigger one, the inequality might not still hold?
    Lets say that f(y)=7, r = 6, and d(x,y)=1, and sup S = 10
    then f(y) ≥ r-d(x,y),
    but we can not say that f(y) ≥ sup S - d(x,y)?
     
    Last edited: Jan 13, 2014
  5. Jan 13, 2014 #4

    jbunniii

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    In general, if we have any bounded set ##S##, and we are given that ##x \geq y## for every ##y \in S##, then we can conclude that ##x \geq \sup S##.

    Proof: suppose that ##x < \sup S##. Now ##\sup S## is the LEAST upper bound of ##S##, so ##x## is not an upper bound of ##S##. Therefore there is some ##y \in S## such that ##x < y##. But this contradicts ## x \geq y## for every ##y \in S##.
     
  6. Jan 13, 2014 #5
    Thanks man, very elegant!
     
  7. Jan 13, 2014 #6

    jbunniii

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    In this example, the inequality ##f(y) \geq r - d(x,y)## is false if ##r > 8##. But ##\sup S = 10##. Therefore there must be some values of ##r \in S## between ##8## and ##10## for which the inequality is false. So this example is different from the situation in your proof.
     
  8. Jan 13, 2014 #7

    jbunniii

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    Here's an even simpler way to put it: the condition ##x \geq y## for all ##y \in S## means exactly that ##x## is an upper bound of ##S##. Therefore ##x## must be at least as big as the LEAST upper bound of ##S##.
     
  9. Jan 13, 2014 #8
    Code (Text):
     
    Ah offcourse, thanks!
     
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