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Math Help - Distance between sets

  1. #1
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    Distance between sets

    Hi, I'm not sure how to prove this statement:

    Prove that if d(A,B)>0 then S = A \cup B is disconnected.

    I'm using this definition of the distance: d(A,b) = inf{|x-y|: x in A, y in B}

    Thanks a lot
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  2. #2
    Junior Member nimon's Avatar
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    If A\cap B\neq\emptyset then it should be easy enough to show that d(A,B) = 0. Can you see a way to do that?
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  3. #3
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    Let \varepsilon  = \frac{{D(A,B)}}{3}.
    Now cover each set: O = \bigcup\limits_{x \in A} {\mathbb{B}(x;\varepsilon )} \;\& \,Q = \bigcup\limits_{x \in B} {\mathbb{B} (x;\varepsilon )} .
    Now you prove that O \cap Q = \emptyset and therefore you have a disconnection.
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  4. #4
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    Quote Originally Posted by nimon View Post
    If A\cap B\neq\emptyset then it should be easy enough to show that d(A,B) = 0. Can you see a way to do that?
    This not helpful.
    Suppose that A = \left( { - \infty ,1} \right]\;\& \;B = \left\{ 1 \right\} \cup \left[ {2,\infty } \right).

    Now quite clearly A \cap B \ne \emptyset \;\& \;D(A,B) = 0.

    BUT A\cup B is not connected.
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  5. #5
    Junior Member nimon's Avatar
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    It is true enough that what I said was unhelpful, but I'm afraid I don't understand your rebuttle. You appear to have contradicted the converse of AKTilted's statement. However I hoped to show that

    A\cap B \neq \emptyset \Rightarrow d(A,B) = 0,

    the contrapositive being:

    d(A,B)>0 \Rightarrow A\cap B = \emptyset

    and therefore that S is disjoint. A proper quashing of that reasoning would be letting A = [0,1) and B = [1,2]. Woops!
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  6. #6
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    What does that have to do with the union being connected or not?
    The OP was about the union being disconnect.
    You did not address that directly.
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