Characterization of interior in metric spaces
$begingroup$
I stumbled upon a theorem in my topology book that didn't have a proof. It says:
For every $x in X$ and $A subset X$,
$x in text{Int} A$ iff $d(x, A^C) > 0$
Here Int denotes interior, A^C is complement of A and d is a metric on X.
If x is in interior of A, then there is an open ball contained in A that contains x. Now, I don't have any idea how to use this to prove that distance between x and XA is positive.
The other direction seems even more difficult.
Any help would be appreciated.
general-topology metric-spaces
asked Nov 25 '18 at 22:05
windircursewindircurse
1,106820
$endgroup$
add a comment |
$begingroup$
I stumbled upon a theorem in my topology book that didn't have a proof. It says:
For every $x in X$ and $A subset X$,
$x in text{Int} A$ iff $d(x, A^C) > 0$
Here Int denotes interior, A^C is complement of A and d is a metric on X.
If x is in interior of A, then there is an open ball contained in A that contains x. Now, I don't have any idea how to use this to prove that distance between x and XA is positive.
The other direction seems even more difficult.
Any help would be appreciated.
general-topology metric-spaces
asked Nov 25 '18 at 22:05
windircursewindircurse
1,106820
$endgroup$
add a comment |
$begingroup$
I stumbled upon a theorem in my topology book that didn't have a proof. It says:
For every $x in X$ and $A subset X$,
$x in text{Int} A$ iff $d(x, A^C) > 0$
Here Int denotes interior, A^C is complement of A and d is a metric on X.
If x is in interior of A, then there is an open ball contained in A that contains x. Now, I don't have any idea how to use this to prove that distance between x and XA is positive.
The other direction seems even more difficult.
Any help would be appreciated.
general-topology metric-spaces
asked Nov 25 '18 at 22:05
windircursewindircurse
1,106820
$endgroup$
I stumbled upon a theorem in my topology book that didn't have a proof. It says:
For every $x in X$ and $A subset X$,
$x in text{Int} A$ iff $d(x, A^C) > 0$
Here Int denotes interior, A^C is complement of A and d is a metric on X.
If x is in interior of A, then there is an open ball contained in A that contains x. Now, I don't have any idea how to use this to prove that distance between x and XA is positive.
The other direction seems even more difficult.
Any help would be appreciated.
general-topology metric-spaces
general-topology metric-spaces
asked Nov 25 '18 at 22:05
windircursewindircurse
1,106820
asked Nov 25 '18 at 22:05
windircursewindircurse
1,106820
asked Nov 25 '18 at 22:05
windircursewindircurse
1,106820
asked Nov 25 '18 at 22:05
windircursewindircurse
1,106820
asked Nov 25 '18 at 22:05
windircursewindircurse
1,106820
1,106820
add a comment |
add a comment |
2 Answers
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If for some $r>0$ we have $B(x,r) subseteq A$,
then let $y$ be any point in $A^complement$.
We know that $d(x,y) ge r$, because otherwise $y in B(x,r)$ and so $y in A$, contradiction, as $y in A^complement$.
This holds for all $y$ in the complement so
$$d(x,A^complement) = inf {d(x,y): y in A^complement }ge r$$
because $r$ is (as I showed) a lower bound for the set of distances and the inf is the greatest lower bound of the set of distances.
So if $x in operatorname{int}(A)$ then $d(x,A^complement) > 0$.
The reverse also holds:
Suppose that $r = d(x,A^complement) > 0$.
Let $y in X$ be such that $d(x,y) < r$ (so $y in B(X,r)$, really). Then $y$ cannot be in $A^complement$ because then $d(x,y)$ would be a number $<r$ on the right in the formula
$$d(x,A^complement) = inf {d(x,z): z in A^complement }$$ and so we'd have $d(x,A^complement) le d(x,y) < r$ which is a contradiction wih the definition of $r$. So $y notin A^complement$ so $y in A$ and we have shown $$B(x,r) subseteq A$$ as required and so $x in operatorname{int}(A)$.
answered Nov 25 '18 at 22:18
Henno BrandsmaHenno Brandsma
106k347114
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add a comment |
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For all $xin text{Int}A$ there is $r>0$ with $B(r,x) subset A$. If $d(x,A^c)=0$ there exists some $yin A^c$ s.t. $d(x,y)<r$: a contradiction. Conversely, if $d(x,y)ge r>0$ for all $yin A^c$ then $B(r,x)subset A$, otherwise, there would be some $yin A^ccap B(r,x)$, which contradicts $d(x,y)ge r$.
answered Nov 25 '18 at 22:18
Guacho PerezGuacho Perez
3,91411132
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
If for some $r>0$ we have $B(x,r) subseteq A$,
then let $y$ be any point in $A^complement$.
We know that $d(x,y) ge r$, because otherwise $y in B(x,r)$ and so $y in A$, contradiction, as $y in A^complement$.
This holds for all $y$ in the complement so
$$d(x,A^complement) = inf {d(x,y): y in A^complement }ge r$$
because $r$ is (as I showed) a lower bound for the set of distances and the inf is the greatest lower bound of the set of distances.
So if $x in operatorname{int}(A)$ then $d(x,A^complement) > 0$.
The reverse also holds:
Suppose that $r = d(x,A^complement) > 0$.
Let $y in X$ be such that $d(x,y) < r$ (so $y in B(X,r)$, really). Then $y$ cannot be in $A^complement$ because then $d(x,y)$ would be a number $<r$ on the right in the formula
$$d(x,A^complement) = inf {d(x,z): z in A^complement }$$ and so we'd have $d(x,A^complement) le d(x,y) < r$ which is a contradiction wih the definition of $r$. So $y notin A^complement$ so $y in A$ and we have shown $$B(x,r) subseteq A$$ as required and so $x in operatorname{int}(A)$.
answered Nov 25 '18 at 22:18
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
If for some $r>0$ we have $B(x,r) subseteq A$,
then let $y$ be any point in $A^complement$.
We know that $d(x,y) ge r$, because otherwise $y in B(x,r)$ and so $y in A$, contradiction, as $y in A^complement$.
This holds for all $y$ in the complement so
$$d(x,A^complement) = inf {d(x,y): y in A^complement }ge r$$
because $r$ is (as I showed) a lower bound for the set of distances and the inf is the greatest lower bound of the set of distances.
So if $x in operatorname{int}(A)$ then $d(x,A^complement) > 0$.
The reverse also holds:
Suppose that $r = d(x,A^complement) > 0$.
Let $y in X$ be such that $d(x,y) < r$ (so $y in B(X,r)$, really). Then $y$ cannot be in $A^complement$ because then $d(x,y)$ would be a number $<r$ on the right in the formula
$$d(x,A^complement) = inf {d(x,z): z in A^complement }$$ and so we'd have $d(x,A^complement) le d(x,y) < r$ which is a contradiction wih the definition of $r$. So $y notin A^complement$ so $y in A$ and we have shown $$B(x,r) subseteq A$$ as required and so $x in operatorname{int}(A)$.
answered Nov 25 '18 at 22:18
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
If for some $r>0$ we have $B(x,r) subseteq A$,
then let $y$ be any point in $A^complement$.
We know that $d(x,y) ge r$, because otherwise $y in B(x,r)$ and so $y in A$, contradiction, as $y in A^complement$.
This holds for all $y$ in the complement so
$$d(x,A^complement) = inf {d(x,y): y in A^complement }ge r$$
because $r$ is (as I showed) a lower bound for the set of distances and the inf is the greatest lower bound of the set of distances.
So if $x in operatorname{int}(A)$ then $d(x,A^complement) > 0$.
The reverse also holds:
Suppose that $r = d(x,A^complement) > 0$.
Let $y in X$ be such that $d(x,y) < r$ (so $y in B(X,r)$, really). Then $y$ cannot be in $A^complement$ because then $d(x,y)$ would be a number $<r$ on the right in the formula
$$d(x,A^complement) = inf {d(x,z): z in A^complement }$$ and so we'd have $d(x,A^complement) le d(x,y) < r$ which is a contradiction wih the definition of $r$. So $y notin A^complement$ so $y in A$ and we have shown $$B(x,r) subseteq A$$ as required and so $x in operatorname{int}(A)$.
answered Nov 25 '18 at 22:18
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
If for some $r>0$ we have $B(x,r) subseteq A$,
then let $y$ be any point in $A^complement$.
We know that $d(x,y) ge r$, because otherwise $y in B(x,r)$ and so $y in A$, contradiction, as $y in A^complement$.
This holds for all $y$ in the complement so
$$d(x,A^complement) = inf {d(x,y): y in A^complement }ge r$$
because $r$ is (as I showed) a lower bound for the set of distances and the inf is the greatest lower bound of the set of distances.
So if $x in operatorname{int}(A)$ then $d(x,A^complement) > 0$.
The reverse also holds:
Suppose that $r = d(x,A^complement) > 0$.
Let $y in X$ be such that $d(x,y) < r$ (so $y in B(X,r)$, really). Then $y$ cannot be in $A^complement$ because then $d(x,y)$ would be a number $<r$ on the right in the formula
$$d(x,A^complement) = inf {d(x,z): z in A^complement }$$ and so we'd have $d(x,A^complement) le d(x,y) < r$ which is a contradiction wih the definition of $r$. So $y notin A^complement$ so $y in A$ and we have shown $$B(x,r) subseteq A$$ as required and so $x in operatorname{int}(A)$.
answered Nov 25 '18 at 22:18
Henno BrandsmaHenno Brandsma
106k347114
answered Nov 25 '18 at 22:18
Henno BrandsmaHenno Brandsma
106k347114
answered Nov 25 '18 at 22:18
Henno BrandsmaHenno Brandsma
106k347114
answered Nov 25 '18 at 22:18
Henno BrandsmaHenno Brandsma
106k347114
106k347114
add a comment |
add a comment |
$begingroup$
For all $xin text{Int}A$ there is $r>0$ with $B(r,x) subset A$. If $d(x,A^c)=0$ there exists some $yin A^c$ s.t. $d(x,y)<r$: a contradiction. Conversely, if $d(x,y)ge r>0$ for all $yin A^c$ then $B(r,x)subset A$, otherwise, there would be some $yin A^ccap B(r,x)$, which contradicts $d(x,y)ge r$.
answered Nov 25 '18 at 22:18
Guacho PerezGuacho Perez
3,91411132
$endgroup$
add a comment |
$begingroup$
For all $xin text{Int}A$ there is $r>0$ with $B(r,x) subset A$. If $d(x,A^c)=0$ there exists some $yin A^c$ s.t. $d(x,y)<r$: a contradiction. Conversely, if $d(x,y)ge r>0$ for all $yin A^c$ then $B(r,x)subset A$, otherwise, there would be some $yin A^ccap B(r,x)$, which contradicts $d(x,y)ge r$.
answered Nov 25 '18 at 22:18
Guacho PerezGuacho Perez
3,91411132
$endgroup$
add a comment |
$begingroup$
For all $xin text{Int}A$ there is $r>0$ with $B(r,x) subset A$. If $d(x,A^c)=0$ there exists some $yin A^c$ s.t. $d(x,y)<r$: a contradiction. Conversely, if $d(x,y)ge r>0$ for all $yin A^c$ then $B(r,x)subset A$, otherwise, there would be some $yin A^ccap B(r,x)$, which contradicts $d(x,y)ge r$.
answered Nov 25 '18 at 22:18
Guacho PerezGuacho Perez
3,91411132
$endgroup$
For all $xin text{Int}A$ there is $r>0$ with $B(r,x) subset A$. If $d(x,A^c)=0$ there exists some $yin A^c$ s.t. $d(x,y)<r$: a contradiction. Conversely, if $d(x,y)ge r>0$ for all $yin A^c$ then $B(r,x)subset A$, otherwise, there would be some $yin A^ccap B(r,x)$, which contradicts $d(x,y)ge r$.
answered Nov 25 '18 at 22:18
Guacho PerezGuacho Perez
3,91411132
answered Nov 25 '18 at 22:18
Guacho PerezGuacho Perez
3,91411132
answered Nov 25 '18 at 22:18
Guacho PerezGuacho Perez
3,91411132
answered Nov 25 '18 at 22:18
Guacho PerezGuacho Perez
3,91411132
3,91411132
add a comment |
add a comment |
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