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Math Help - Real Numbers - Real Anaylsis

  1. #1
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    Real Numbers - Real Anaylsis

    Show that if a,b \in \Re, and a \neq b, then there exist \epsilon-neighborhoods U of a and V of b such that U \cap V = \emptyset
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  2. #2
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    Let \varepsilon  = \frac{{\left| {a - b} \right|}}{4}.
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    any reason why you divided by 4?
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    Quote Originally Posted by hockey777 View Post
    any reason why you divided by 4?
    Two reasons: it works & I like 4.
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  5. #5
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    Sorry one more question. What else was throwing me off in this problem was the fact I don't think it's clear for a beginner like me. I think the problem suggest that \epsilon should be the same for both, but I'm not sure. Is that true?
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  6. #6
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    No there is no requirement the neighborhoods have the same radius.
    But I gave you a value that will work for both.

    Let U = \left\{ {x:\left| {a - x} \right| < \varepsilon } \right\}\,\& \,V = \left\{ {x:\left| {b - x} \right| < \varepsilon } \right\}.

    Now suppose that y \in U \cap V \Rightarrow \quad \left| {a - b} \right| \le \left| {a - y} \right| + \left| {y - b} \right| < \varepsilon  + \varepsilon  = \frac{{\left| {a - b} \right|}}{2}.
    That is a clear contradiction.
    So U & V must have nothing in common.
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