Prove that $U={x in Xmid d(x,A)<d(x,B)}$ is open when $A$ and $B$ are disjoint
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In a metric space $(X,d)$, I have the set $$U={x in Xmid d(x,A)<d(x,B)}$$
where $A$ and $B$ are disjoint subsets. I need to show that $U$ is open in $(X,d)$. I tried taking the radius of an open ball centre $xin U$ to be less than $d(x,C)$ where $C={xin Xmid d(x,A)=d(x,C)}$ but I could not get anywhere. Is this the right strategy? I would really appreciate help, thank you!
metric-spaces
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$begingroup$
In a metric space $(X,d)$, I have the set $$U={x in Xmid d(x,A)<d(x,B)}$$
where $A$ and $B$ are disjoint subsets. I need to show that $U$ is open in $(X,d)$. I tried taking the radius of an open ball centre $xin U$ to be less than $d(x,C)$ where $C={xin Xmid d(x,A)=d(x,C)}$ but I could not get anywhere. Is this the right strategy? I would really appreciate help, thank you!
metric-spaces
$endgroup$
add a comment |
$begingroup$
In a metric space $(X,d)$, I have the set $$U={x in Xmid d(x,A)<d(x,B)}$$
where $A$ and $B$ are disjoint subsets. I need to show that $U$ is open in $(X,d)$. I tried taking the radius of an open ball centre $xin U$ to be less than $d(x,C)$ where $C={xin Xmid d(x,A)=d(x,C)}$ but I could not get anywhere. Is this the right strategy? I would really appreciate help, thank you!
metric-spaces
$endgroup$
In a metric space $(X,d)$, I have the set $$U={x in Xmid d(x,A)<d(x,B)}$$
where $A$ and $B$ are disjoint subsets. I need to show that $U$ is open in $(X,d)$. I tried taking the radius of an open ball centre $xin U$ to be less than $d(x,C)$ where $C={xin Xmid d(x,A)=d(x,C)}$ but I could not get anywhere. Is this the right strategy? I would really appreciate help, thank you!
metric-spaces
metric-spaces
edited 6 hours ago
Asaf Karagila♦
306k33438769
306k33438769
asked 13 hours ago
Sean ThrasherSean Thrasher
384
384
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2 Answers
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$begingroup$
$f(x)=d(x,B)-d(x,A)$ is continuous and $U=f^{-1}(x:x>0)$
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$begingroup$
Your strategy will not work in an arbitrary metric space. Consider for example
$ X = mathbb R setminus{0}$ with the usual distance and then $A={-1}$, $B={1}$. Then your $C$ is empty.
Instead, consider something like the ball around $xin U$ of radius $frac12(d(x,B)-d(x,A))$.
(Tsemo Aristide's suggestion is slicker than this, if you already know that $d(x,A)$ is a continuous function of $x$ and that preimages of open sets under continuous functions are open).
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2 Answers
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2 Answers
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$begingroup$
$f(x)=d(x,B)-d(x,A)$ is continuous and $U=f^{-1}(x:x>0)$
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add a comment |
$begingroup$
$f(x)=d(x,B)-d(x,A)$ is continuous and $U=f^{-1}(x:x>0)$
$endgroup$
add a comment |
$begingroup$
$f(x)=d(x,B)-d(x,A)$ is continuous and $U=f^{-1}(x:x>0)$
$endgroup$
$f(x)=d(x,B)-d(x,A)$ is continuous and $U=f^{-1}(x:x>0)$
answered 12 hours ago
Tsemo AristideTsemo Aristide
59.4k11446
59.4k11446
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$begingroup$
Your strategy will not work in an arbitrary metric space. Consider for example
$ X = mathbb R setminus{0}$ with the usual distance and then $A={-1}$, $B={1}$. Then your $C$ is empty.
Instead, consider something like the ball around $xin U$ of radius $frac12(d(x,B)-d(x,A))$.
(Tsemo Aristide's suggestion is slicker than this, if you already know that $d(x,A)$ is a continuous function of $x$ and that preimages of open sets under continuous functions are open).
$endgroup$
add a comment |
$begingroup$
Your strategy will not work in an arbitrary metric space. Consider for example
$ X = mathbb R setminus{0}$ with the usual distance and then $A={-1}$, $B={1}$. Then your $C$ is empty.
Instead, consider something like the ball around $xin U$ of radius $frac12(d(x,B)-d(x,A))$.
(Tsemo Aristide's suggestion is slicker than this, if you already know that $d(x,A)$ is a continuous function of $x$ and that preimages of open sets under continuous functions are open).
$endgroup$
add a comment |
$begingroup$
Your strategy will not work in an arbitrary metric space. Consider for example
$ X = mathbb R setminus{0}$ with the usual distance and then $A={-1}$, $B={1}$. Then your $C$ is empty.
Instead, consider something like the ball around $xin U$ of radius $frac12(d(x,B)-d(x,A))$.
(Tsemo Aristide's suggestion is slicker than this, if you already know that $d(x,A)$ is a continuous function of $x$ and that preimages of open sets under continuous functions are open).
$endgroup$
Your strategy will not work in an arbitrary metric space. Consider for example
$ X = mathbb R setminus{0}$ with the usual distance and then $A={-1}$, $B={1}$. Then your $C$ is empty.
Instead, consider something like the ball around $xin U$ of radius $frac12(d(x,B)-d(x,A))$.
(Tsemo Aristide's suggestion is slicker than this, if you already know that $d(x,A)$ is a continuous function of $x$ and that preimages of open sets under continuous functions are open).
edited 12 hours ago
answered 12 hours ago
Henning MakholmHenning Makholm
242k17308549
242k17308549
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