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...
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
asked Dec 3 '18 at 0:39
Marco Gutiérrez
153
add a comment |
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
asked Dec 3 '18 at 0:39
Marco Gutiérrez
153
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.
– NL1992
Dec 3 '18 at 0:51
$$Y={xinmathbb R^nmid d(x,A)<d(x,B)}qquad Z={xinmathbb R^nmid d(x,A)>d(x,B)}$$
– Did
Dec 3 '18 at 1:37
Did, your answer was right. Thanks for the comment
– Marco Gutiérrez
Dec 3 '18 at 2:57
add a comment |
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
asked Dec 3 '18 at 0:39
Marco Gutiérrez
153
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
general-topology analysis covering-spaces
asked Dec 3 '18 at 0:39
Marco Gutiérrez
153
asked Dec 3 '18 at 0:39
Marco Gutiérrez
153
edited Dec 3 '18 at 2:51
asked Dec 3 '18 at 0:39
Marco Gutiérrez
153
asked Dec 3 '18 at 0:39
Marco Gutiérrez
153
asked Dec 3 '18 at 0:39
Marco Gutiérrez
153
153
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.
– NL1992
Dec 3 '18 at 0:51
$$Y={xinmathbb R^nmid d(x,A)<d(x,B)}qquad Z={xinmathbb R^nmid d(x,A)>d(x,B)}$$
– Did
Dec 3 '18 at 1:37
Did, your answer was right. Thanks for the comment
– Marco Gutiérrez
Dec 3 '18 at 2:57
add a comment |
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.
– NL1992
Dec 3 '18 at 0:51
$$Y={xinmathbb R^nmid d(x,A)<d(x,B)}qquad Z={xinmathbb R^nmid d(x,A)>d(x,B)}$$
– Did
Dec 3 '18 at 1:37
Did, your answer was right. Thanks for the comment
– Marco Gutiérrez
Dec 3 '18 at 2:57
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.
– NL1992
Dec 3 '18 at 0:51
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.
– NL1992
Dec 3 '18 at 0:51
$$Y={xinmathbb R^nmid d(x,A)<d(x,B)}qquad Z={xinmathbb R^nmid d(x,A)>d(x,B)}$$
– Did
Dec 3 '18 at 1:37
$$Y={xinmathbb R^nmid d(x,A)<d(x,B)}qquad Z={xinmathbb R^nmid d(x,A)>d(x,B)}$$
– Did
Dec 3 '18 at 1:37
Did, your answer was right. Thanks for the comment
– Marco Gutiérrez
Dec 3 '18 at 2:57
Did, your answer was right. Thanks for the comment
– Marco Gutiérrez
Dec 3 '18 at 2:57
add a comment |
1 Answer
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See Did's comment. There was an error in my previous solution so I deleted it.
answered Dec 3 '18 at 2:30
simplemath
363
add a comment |
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See Did's comment. There was an error in my previous solution so I deleted it.
answered Dec 3 '18 at 2:30
simplemath
363
add a comment |
See Did's comment. There was an error in my previous solution so I deleted it.
answered Dec 3 '18 at 2:30
simplemath
363
add a comment |
See Did's comment. There was an error in my previous solution so I deleted it.
answered Dec 3 '18 at 2:30
simplemath
363
See Did's comment. There was an error in my previous solution so I deleted it.
answered Dec 3 '18 at 2:30
simplemath
363
edited Dec 3 '18 at 7:19
answered Dec 3 '18 at 2:30
simplemath
363
answered Dec 3 '18 at 2:30
simplemath
363
answered Dec 3 '18 at 2:30
simplemath
363
363
add a comment |
add a comment |
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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.
– NL1992
Dec 3 '18 at 0:51
$$Y={xinmathbb R^nmid d(x,A)<d(x,B)}qquad Z={xinmathbb R^nmid d(x,A)>d(x,B)}$$
– Did
Dec 3 '18 at 1:37
Did, your answer was right. Thanks for the comment
– Marco Gutiérrez
Dec 3 '18 at 2:57