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Math Help - Neighborhoods

  1. #1
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    Neighborhoods

    Let x and y be distinct real numbers. Prove that there is a nbhood P of x and a nbhood Q of y such that PintersectQ=null set.
    So, we have (x-epsilon, x+epsilon) and (y-epsilon, y+epsilon).
    I have an idea of how to do this but I'm getting stuck. So, I think I need to define epsilon, right because these are two different epsilons. This is where I get stuck. Can someone give me pointers on defining epsilon?
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  2. #2
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    Quote Originally Posted by kathrynmath View Post
    So, we have (x-epsilon, x+epsilon) and (y-epsilon, y+epsilon).
    Let \epsilon = |x-y|/2.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Let \epsilon = |x-y|/2.
    How did you even get that?
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  4. #4
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    Quote Originally Posted by kathrynmath View Post
    How did you even get that?
    Draw a picture. Draw a real number line and put x and y. The distance between those points is |x-y|. Thus, if you draw a interval around them half the length |x-y|/2. Then those two intervals would not intersect.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    Draw a picture. Draw a real number line and put x and y. The distance between those points is |x-y|. Thus, if you draw a interval around them half the length |x-y|/2. Then those two intervals would not intersect.
    Ok, thanks. I get that.
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  6. #6
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    I'm not quite sure how to prove it,though. I understand why it works and all, but I know a picture doesn't qualify as a proof.
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  7. #7
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    Suppose that \varepsilon  = \frac{{\left| {x - y} \right|}}{2} > 0 and that z \in N_\varepsilon  (x) \cap N_\varepsilon  (y).
    \left| {x - y} \right| \leqslant \left| {x - z} \right| + \left| {z - y} \right| < \varepsilon  + \varepsilon  = \left| {x - y} \right|
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  8. #8
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    Quote Originally Posted by Plato View Post
    Suppose that \varepsilon = \frac{{\left| {x - y} \right|}}{2} > 0 and that z \in N_\varepsilon (x) \cap N_\varepsilon (y).
    \left| {x - y} \right| \leqslant \left| {x - z} \right| + \left| {z - y} \right| < \varepsilon + \varepsilon = \left| {x - y} \right|

    Ok, I think I understand this. So, is N the neighborhood?
    Is this like a triangle inequality?
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  9. #9
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    {\color{blue}\left| {x - y} \right|} \leqslant \left| {x - z} \right| + \left| {z - y} \right| {\color{blue}<} \varepsilon  + \varepsilon  = {\color{blue}\left| {x - y} \right|}
    Do you see the contradiction in the blues?
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  10. #10
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    Quote Originally Posted by Plato View Post
    {\color{blue}\left| {x - y} \right|} \leqslant \left| {x - z} \right| + \left| {z - y} \right| {\color{blue}<} \varepsilon + \varepsilon = {\color{blue}\left| {x - y} \right|}
    Do you see the contradiction in the blues?
    I'm trying to see the contradiction. Why is < epsilon a contradiciton? I'm not getting it....
    and the x-y in the end appears correct because 2epsilon = x-y
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  11. #11
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    Quote Originally Posted by kathrynmath View Post
    I'm trying to see the contradiction. Why is < epsilon a contradiciton? I'm not getting it....
    and the x-y in the end appears correct because 2epsilon = x-y
    A number cannot be less than itself.
    There in blue you see \left| {x - y} \right| < \left| {x - y} \right|.
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  12. #12
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    Quote Originally Posted by Plato View Post
    A number cannot be less than itself.
    There in blue you see \left| {x - y} \right| < \left| {x - y} \right|.
    Oh ok. So then this is a a contradicition and leads to the fact that P intersect Q = null set?
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