If i have two closed and disjoint sets A and B in a set X c $R^{m}$, are there any open and disjoint sets in...












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I have a set $X$ $c$ $R^{m}$ and a pair of two closed and disjoint sets in X, called A and B. What i am trying to found at least a pair of sets Y and Z that are open and disjoint in a set X c $R^{m}$ that include A and B, respectively.



For A, i tried using a cover made of all the balls with center in a point of A and an arbitrary ratio intersected with X - A, but if i do the same for B, how could i know if those two covers are disjoint?










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  • $begingroup$
    A nice property of metrisable spaces such as $Bbb R^n$ is that they have a countable basis. You can use that for your proof and that way 'disjoin' adjacent open sets.
    $endgroup$
    – NL1992
    Dec 3 '18 at 0:51












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    $$Y={xinmathbb R^nmid d(x,A)<d(x,B)}qquad Z={xinmathbb R^nmid d(x,A)>d(x,B)}$$
    $endgroup$
    – Did
    Dec 3 '18 at 1:37










  • $begingroup$
    Did, your answer was right. Thanks for the comment
    $endgroup$
    – Marco Gutiérrez
    Dec 3 '18 at 2:57
















1












$begingroup$


I have a set $X$ $c$ $R^{m}$ and a pair of two closed and disjoint sets in X, called A and B. What i am trying to found at least a pair of sets Y and Z that are open and disjoint in a set X c $R^{m}$ that include A and B, respectively.



For A, i tried using a cover made of all the balls with center in a point of A and an arbitrary ratio intersected with X - A, but if i do the same for B, how could i know if those two covers are disjoint?










share|cite|improve this question











$endgroup$












  • $begingroup$
    A nice property of metrisable spaces such as $Bbb R^n$ is that they have a countable basis. You can use that for your proof and that way 'disjoin' adjacent open sets.
    $endgroup$
    – NL1992
    Dec 3 '18 at 0:51












  • $begingroup$
    $$Y={xinmathbb R^nmid d(x,A)<d(x,B)}qquad Z={xinmathbb R^nmid d(x,A)>d(x,B)}$$
    $endgroup$
    – Did
    Dec 3 '18 at 1:37










  • $begingroup$
    Did, your answer was right. Thanks for the comment
    $endgroup$
    – Marco Gutiérrez
    Dec 3 '18 at 2:57














1












1








1





$begingroup$


I have a set $X$ $c$ $R^{m}$ and a pair of two closed and disjoint sets in X, called A and B. What i am trying to found at least a pair of sets Y and Z that are open and disjoint in a set X c $R^{m}$ that include A and B, respectively.



For A, i tried using a cover made of all the balls with center in a point of A and an arbitrary ratio intersected with X - A, but if i do the same for B, how could i know if those two covers are disjoint?










share|cite|improve this question











$endgroup$




I have a set $X$ $c$ $R^{m}$ and a pair of two closed and disjoint sets in X, called A and B. What i am trying to found at least a pair of sets Y and Z that are open and disjoint in a set X c $R^{m}$ that include A and B, respectively.



For A, i tried using a cover made of all the balls with center in a point of A and an arbitrary ratio intersected with X - A, but if i do the same for B, how could i know if those two covers are disjoint?







general-topology analysis covering-spaces






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edited Dec 3 '18 at 2:51







Marco Gutiérrez

















asked Dec 3 '18 at 0:39









Marco GutiérrezMarco Gutiérrez

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153












  • $begingroup$
    A nice property of metrisable spaces such as $Bbb R^n$ is that they have a countable basis. You can use that for your proof and that way 'disjoin' adjacent open sets.
    $endgroup$
    – NL1992
    Dec 3 '18 at 0:51












  • $begingroup$
    $$Y={xinmathbb R^nmid d(x,A)<d(x,B)}qquad Z={xinmathbb R^nmid d(x,A)>d(x,B)}$$
    $endgroup$
    – Did
    Dec 3 '18 at 1:37










  • $begingroup$
    Did, your answer was right. Thanks for the comment
    $endgroup$
    – Marco Gutiérrez
    Dec 3 '18 at 2:57


















  • $begingroup$
    A nice property of metrisable spaces such as $Bbb R^n$ is that they have a countable basis. You can use that for your proof and that way 'disjoin' adjacent open sets.
    $endgroup$
    – NL1992
    Dec 3 '18 at 0:51












  • $begingroup$
    $$Y={xinmathbb R^nmid d(x,A)<d(x,B)}qquad Z={xinmathbb R^nmid d(x,A)>d(x,B)}$$
    $endgroup$
    – Did
    Dec 3 '18 at 1:37










  • $begingroup$
    Did, your answer was right. Thanks for the comment
    $endgroup$
    – Marco Gutiérrez
    Dec 3 '18 at 2:57
















$begingroup$
A nice property of metrisable spaces such as $Bbb R^n$ is that they have a countable basis. You can use that for your proof and that way 'disjoin' adjacent open sets.
$endgroup$
– NL1992
Dec 3 '18 at 0:51






$begingroup$
A nice property of metrisable spaces such as $Bbb R^n$ is that they have a countable basis. You can use that for your proof and that way 'disjoin' adjacent open sets.
$endgroup$
– NL1992
Dec 3 '18 at 0:51














$begingroup$
$$Y={xinmathbb R^nmid d(x,A)<d(x,B)}qquad Z={xinmathbb R^nmid d(x,A)>d(x,B)}$$
$endgroup$
– Did
Dec 3 '18 at 1:37




$begingroup$
$$Y={xinmathbb R^nmid d(x,A)<d(x,B)}qquad Z={xinmathbb R^nmid d(x,A)>d(x,B)}$$
$endgroup$
– Did
Dec 3 '18 at 1:37












$begingroup$
Did, your answer was right. Thanks for the comment
$endgroup$
– Marco Gutiérrez
Dec 3 '18 at 2:57




$begingroup$
Did, your answer was right. Thanks for the comment
$endgroup$
– Marco Gutiérrez
Dec 3 '18 at 2:57










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See Did's comment. There was an error in my previous solution so I deleted it.






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    $begingroup$

    See Did's comment. There was an error in my previous solution so I deleted it.






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      $begingroup$

      See Did's comment. There was an error in my previous solution so I deleted it.






      share|cite|improve this answer











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        $begingroup$

        See Did's comment. There was an error in my previous solution so I deleted it.






        share|cite|improve this answer











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        See Did's comment. There was an error in my previous solution so I deleted it.







        share|cite|improve this answer














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        edited Dec 3 '18 at 7:19

























        answered Dec 3 '18 at 2:30









        simplemathsimplemath

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