If distance between the two sets is zero then their intersection is empty?
up vote
3
down vote
favorite
Let $(X,d)$ be a metric space and $A$ and $B$ two non-empty sets in
$X$.
If $d(A,B)= 0$ then, is their intersection empty?
I attempted this problem writing:
$d(A, B)=infleft{ d(x,y): xin A, yin Bright}=0 implies d(x,y)= 0 implies x=y$
That is, if $d(A,B) = 0$ then there exists $x in A cap B$, so the intersection is not empty!
My question is whether I am taking the infimum wrong or doing anything else wrongly? How can I show this with an example?
proof-verification metric-spaces
edited 22 hours ago
Javier
1,95521131
asked Sep 30 '15 at 13:03
Kavita
499517
add a comment |
up vote
3
down vote
favorite
Let $(X,d)$ be a metric space and $A$ and $B$ two non-empty sets in
$X$.
If $d(A,B)= 0$ then, is their intersection empty?
I attempted this problem writing:
$d(A, B)=infleft{ d(x,y): xin A, yin Bright}=0 implies d(x,y)= 0 implies x=y$
That is, if $d(A,B) = 0$ then there exists $x in A cap B$, so the intersection is not empty!
My question is whether I am taking the infimum wrong or doing anything else wrongly? How can I show this with an example?
proof-verification metric-spaces
edited 22 hours ago
Javier
1,95521131
asked Sep 30 '15 at 13:03
Kavita
499517
6
Well, the intersection could be empty, but it need not be. You seem to be assuming that the infimum is attained inside the set, which it need not (consider for example the disjoint intervals $(0,1)$ and $(1,2)$ on the real like).
– Tobias Kildetoft
Sep 30 '15 at 13:06
1
So is there no final conclusion about their intersection?
– Kavita
Sep 30 '15 at 13:14
1
No, your conclusion that $Acap B$ is non-empty is wrong. What you can say is that there exists a point $c$ such that $c$ is a limit point in both $A$ and $B$. For example $mathbb R^+$ and $mathbb R^-$ have distance $0$, but there is no real number that is both positive and negative (but a limit point $0$ exists to both sets).
– skyking
Sep 30 '15 at 13:28
@skyking Not even that, for the limit point can be "missing" from $X$. Think of $A = {(x_1, 0) mid x_1 in mathbf R}$, and $B = {x mid x_1 x_2 = 1}$ in $X = mathbf R^2$.
– martini
Sep 30 '15 at 13:30
@martini My bad, there's a sequence in $A$ and $B$ such that $lim d(a_j, b_j) = 0$. You need $A$ and $B$ to be part of a compact set to get a limit point (unless I'm mistaken again).
– skyking
Sep 30 '15 at 13:46
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $(X,d)$ be a metric space and $A$ and $B$ two non-empty sets in
$X$.
If $d(A,B)= 0$ then, is their intersection empty?
I attempted this problem writing:
$d(A, B)=infleft{ d(x,y): xin A, yin Bright}=0 implies d(x,y)= 0 implies x=y$
That is, if $d(A,B) = 0$ then there exists $x in A cap B$, so the intersection is not empty!
My question is whether I am taking the infimum wrong or doing anything else wrongly? How can I show this with an example?
proof-verification metric-spaces
edited 22 hours ago
Javier
1,95521131
asked Sep 30 '15 at 13:03
Kavita
499517
Let $(X,d)$ be a metric space and $A$ and $B$ two non-empty sets in
$X$.
If $d(A,B)= 0$ then, is their intersection empty?
I attempted this problem writing:
$d(A, B)=infleft{ d(x,y): xin A, yin Bright}=0 implies d(x,y)= 0 implies x=y$
That is, if $d(A,B) = 0$ then there exists $x in A cap B$, so the intersection is not empty!
My question is whether I am taking the infimum wrong or doing anything else wrongly? How can I show this with an example?
proof-verification metric-spaces
proof-verification metric-spaces
edited 22 hours ago
Javier
1,95521131
asked Sep 30 '15 at 13:03
Kavita
499517
edited 22 hours ago
Javier
1,95521131
asked Sep 30 '15 at 13:03
Kavita
499517
edited 22 hours ago
Javier
1,95521131
edited 22 hours ago
Javier
1,95521131
edited 22 hours ago
Javier
1,95521131
1,95521131
asked Sep 30 '15 at 13:03
Kavita
499517
asked Sep 30 '15 at 13:03
Kavita
499517
asked Sep 30 '15 at 13:03
Kavita
499517
499517
6
Well, the intersection could be empty, but it need not be. You seem to be assuming that the infimum is attained inside the set, which it need not (consider for example the disjoint intervals $(0,1)$ and $(1,2)$ on the real like).
– Tobias Kildetoft
Sep 30 '15 at 13:06
1
So is there no final conclusion about their intersection?
– Kavita
Sep 30 '15 at 13:14
1
No, your conclusion that $Acap B$ is non-empty is wrong. What you can say is that there exists a point $c$ such that $c$ is a limit point in both $A$ and $B$. For example $mathbb R^+$ and $mathbb R^-$ have distance $0$, but there is no real number that is both positive and negative (but a limit point $0$ exists to both sets).
– skyking
Sep 30 '15 at 13:28
@skyking Not even that, for the limit point can be "missing" from $X$. Think of $A = {(x_1, 0) mid x_1 in mathbf R}$, and $B = {x mid x_1 x_2 = 1}$ in $X = mathbf R^2$.
– martini
Sep 30 '15 at 13:30
@martini My bad, there's a sequence in $A$ and $B$ such that $lim d(a_j, b_j) = 0$. You need $A$ and $B$ to be part of a compact set to get a limit point (unless I'm mistaken again).
– skyking
Sep 30 '15 at 13:46
add a comment |
6
Well, the intersection could be empty, but it need not be. You seem to be assuming that the infimum is attained inside the set, which it need not (consider for example the disjoint intervals $(0,1)$ and $(1,2)$ on the real like).
– Tobias Kildetoft
Sep 30 '15 at 13:06
1
So is there no final conclusion about their intersection?
– Kavita
Sep 30 '15 at 13:14
1
No, your conclusion that $Acap B$ is non-empty is wrong. What you can say is that there exists a point $c$ such that $c$ is a limit point in both $A$ and $B$. For example $mathbb R^+$ and $mathbb R^-$ have distance $0$, but there is no real number that is both positive and negative (but a limit point $0$ exists to both sets).
– skyking
Sep 30 '15 at 13:28
@skyking Not even that, for the limit point can be "missing" from $X$. Think of $A = {(x_1, 0) mid x_1 in mathbf R}$, and $B = {x mid x_1 x_2 = 1}$ in $X = mathbf R^2$.
– martini
Sep 30 '15 at 13:30
@martini My bad, there's a sequence in $A$ and $B$ such that $lim d(a_j, b_j) = 0$. You need $A$ and $B$ to be part of a compact set to get a limit point (unless I'm mistaken again).
– skyking
Sep 30 '15 at 13:46
6
6
Well, the intersection could be empty, but it need not be. You seem to be assuming that the infimum is attained inside the set, which it need not (consider for example the disjoint intervals $(0,1)$ and $(1,2)$ on the real like).
– Tobias Kildetoft
Sep 30 '15 at 13:06
Well, the intersection could be empty, but it need not be. You seem to be assuming that the infimum is attained inside the set, which it need not (consider for example the disjoint intervals $(0,1)$ and $(1,2)$ on the real like).
– Tobias Kildetoft
Sep 30 '15 at 13:06
1
1
So is there no final conclusion about their intersection?
– Kavita
Sep 30 '15 at 13:14
So is there no final conclusion about their intersection?
– Kavita
Sep 30 '15 at 13:14
1
1
No, your conclusion that $Acap B$ is non-empty is wrong. What you can say is that there exists a point $c$ such that $c$ is a limit point in both $A$ and $B$. For example $mathbb R^+$ and $mathbb R^-$ have distance $0$, but there is no real number that is both positive and negative (but a limit point $0$ exists to both sets).
– skyking
Sep 30 '15 at 13:28
No, your conclusion that $Acap B$ is non-empty is wrong. What you can say is that there exists a point $c$ such that $c$ is a limit point in both $A$ and $B$. For example $mathbb R^+$ and $mathbb R^-$ have distance $0$, but there is no real number that is both positive and negative (but a limit point $0$ exists to both sets).
– skyking
Sep 30 '15 at 13:28
@skyking Not even that, for the limit point can be "missing" from $X$. Think of $A = {(x_1, 0) mid x_1 in mathbf R}$, and $B = {x mid x_1 x_2 = 1}$ in $X = mathbf R^2$.
– martini
Sep 30 '15 at 13:30
@skyking Not even that, for the limit point can be "missing" from $X$. Think of $A = {(x_1, 0) mid x_1 in mathbf R}$, and $B = {x mid x_1 x_2 = 1}$ in $X = mathbf R^2$.
– martini
Sep 30 '15 at 13:30
@martini My bad, there's a sequence in $A$ and $B$ such that $lim d(a_j, b_j) = 0$. You need $A$ and $B$ to be part of a compact set to get a limit point (unless I'm mistaken again).
– skyking
Sep 30 '15 at 13:46
@martini My bad, there's a sequence in $A$ and $B$ such that $lim d(a_j, b_j) = 0$. You need $A$ and $B$ to be part of a compact set to get a limit point (unless I'm mistaken again).
– skyking
Sep 30 '15 at 13:46
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Without any further assumption on $A$ and $B$, there is no conclusion about their intersection:
In $X = defR{mathbf R}R$ with the euclidian metric, $A = [0,1]$ and $B = (1,2)$ have $d(A,B) = 0$ and empty intersection, but $A$ and $B' = [1,2]$ have also $d(A, B') = 0$.
Even if we suppose $A$ and $B$ to be closed, we can have $d(A,B) = 0$ with empty intersection. Consider $A = R times {0}$ and $B= {x in R^2 mid x_1 x_2 = 1}$. Here $d(A,B) = 0$ due to $d(a_n, b_n) = frac 1n$ for $a_n = (n,0) in A$ and $b_n = (n, frac 1n) in B$.
If we suppose $A$ to be compact and $B$ to be closed, something can be said. It that case, we have that $d(A,B) = 0$ implies $A cap B ne emptyset$. For choose $a_n in A$ and $b_n in B$ such that $d(a_n, b_n) to d(A,B)=0$. As $A$ is compact $(a_n)$ has a convergent subsequence, we may suppose w. l. o. g. that $a_n to a in A$. As $d(a_n, b_n) to 0$, we have that
$$ d(b_n, a) le d(a_n, b_n) + d(a_n,a) to 0 $$
that is $b_n to a$. As $B$ is closed, $a in B$, that is $ain A cap B$.
answered Sep 30 '15 at 13:26
martini
69.9k45889
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Without any further assumption on $A$ and $B$, there is no conclusion about their intersection:
In $X = defR{mathbf R}R$ with the euclidian metric, $A = [0,1]$ and $B = (1,2)$ have $d(A,B) = 0$ and empty intersection, but $A$ and $B' = [1,2]$ have also $d(A, B') = 0$.
Even if we suppose $A$ and $B$ to be closed, we can have $d(A,B) = 0$ with empty intersection. Consider $A = R times {0}$ and $B= {x in R^2 mid x_1 x_2 = 1}$. Here $d(A,B) = 0$ due to $d(a_n, b_n) = frac 1n$ for $a_n = (n,0) in A$ and $b_n = (n, frac 1n) in B$.
If we suppose $A$ to be compact and $B$ to be closed, something can be said. It that case, we have that $d(A,B) = 0$ implies $A cap B ne emptyset$. For choose $a_n in A$ and $b_n in B$ such that $d(a_n, b_n) to d(A,B)=0$. As $A$ is compact $(a_n)$ has a convergent subsequence, we may suppose w. l. o. g. that $a_n to a in A$. As $d(a_n, b_n) to 0$, we have that
$$ d(b_n, a) le d(a_n, b_n) + d(a_n,a) to 0 $$
that is $b_n to a$. As $B$ is closed, $a in B$, that is $ain A cap B$.
answered Sep 30 '15 at 13:26
martini
69.9k45889
add a comment |
up vote
2
down vote
accepted
Without any further assumption on $A$ and $B$, there is no conclusion about their intersection:
In $X = defR{mathbf R}R$ with the euclidian metric, $A = [0,1]$ and $B = (1,2)$ have $d(A,B) = 0$ and empty intersection, but $A$ and $B' = [1,2]$ have also $d(A, B') = 0$.
Even if we suppose $A$ and $B$ to be closed, we can have $d(A,B) = 0$ with empty intersection. Consider $A = R times {0}$ and $B= {x in R^2 mid x_1 x_2 = 1}$. Here $d(A,B) = 0$ due to $d(a_n, b_n) = frac 1n$ for $a_n = (n,0) in A$ and $b_n = (n, frac 1n) in B$.
If we suppose $A$ to be compact and $B$ to be closed, something can be said. It that case, we have that $d(A,B) = 0$ implies $A cap B ne emptyset$. For choose $a_n in A$ and $b_n in B$ such that $d(a_n, b_n) to d(A,B)=0$. As $A$ is compact $(a_n)$ has a convergent subsequence, we may suppose w. l. o. g. that $a_n to a in A$. As $d(a_n, b_n) to 0$, we have that
$$ d(b_n, a) le d(a_n, b_n) + d(a_n,a) to 0 $$
that is $b_n to a$. As $B$ is closed, $a in B$, that is $ain A cap B$.
answered Sep 30 '15 at 13:26
martini
69.9k45889
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Without any further assumption on $A$ and $B$, there is no conclusion about their intersection:
In $X = defR{mathbf R}R$ with the euclidian metric, $A = [0,1]$ and $B = (1,2)$ have $d(A,B) = 0$ and empty intersection, but $A$ and $B' = [1,2]$ have also $d(A, B') = 0$.
Even if we suppose $A$ and $B$ to be closed, we can have $d(A,B) = 0$ with empty intersection. Consider $A = R times {0}$ and $B= {x in R^2 mid x_1 x_2 = 1}$. Here $d(A,B) = 0$ due to $d(a_n, b_n) = frac 1n$ for $a_n = (n,0) in A$ and $b_n = (n, frac 1n) in B$.
If we suppose $A$ to be compact and $B$ to be closed, something can be said. It that case, we have that $d(A,B) = 0$ implies $A cap B ne emptyset$. For choose $a_n in A$ and $b_n in B$ such that $d(a_n, b_n) to d(A,B)=0$. As $A$ is compact $(a_n)$ has a convergent subsequence, we may suppose w. l. o. g. that $a_n to a in A$. As $d(a_n, b_n) to 0$, we have that
$$ d(b_n, a) le d(a_n, b_n) + d(a_n,a) to 0 $$
that is $b_n to a$. As $B$ is closed, $a in B$, that is $ain A cap B$.
answered Sep 30 '15 at 13:26
martini
69.9k45889
Without any further assumption on $A$ and $B$, there is no conclusion about their intersection:
In $X = defR{mathbf R}R$ with the euclidian metric, $A = [0,1]$ and $B = (1,2)$ have $d(A,B) = 0$ and empty intersection, but $A$ and $B' = [1,2]$ have also $d(A, B') = 0$.
Even if we suppose $A$ and $B$ to be closed, we can have $d(A,B) = 0$ with empty intersection. Consider $A = R times {0}$ and $B= {x in R^2 mid x_1 x_2 = 1}$. Here $d(A,B) = 0$ due to $d(a_n, b_n) = frac 1n$ for $a_n = (n,0) in A$ and $b_n = (n, frac 1n) in B$.
If we suppose $A$ to be compact and $B$ to be closed, something can be said. It that case, we have that $d(A,B) = 0$ implies $A cap B ne emptyset$. For choose $a_n in A$ and $b_n in B$ such that $d(a_n, b_n) to d(A,B)=0$. As $A$ is compact $(a_n)$ has a convergent subsequence, we may suppose w. l. o. g. that $a_n to a in A$. As $d(a_n, b_n) to 0$, we have that
$$ d(b_n, a) le d(a_n, b_n) + d(a_n,a) to 0 $$
that is $b_n to a$. As $B$ is closed, $a in B$, that is $ain A cap B$.
answered Sep 30 '15 at 13:26
martini
69.9k45889
answered Sep 30 '15 at 13:26
martini
69.9k45889
answered Sep 30 '15 at 13:26
martini
69.9k45889
answered Sep 30 '15 at 13:26
martini
69.9k45889
69.9k45889
add a comment |
add a comment |
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6
Well, the intersection could be empty, but it need not be. You seem to be assuming that the infimum is attained inside the set, which it need not (consider for example the disjoint intervals $(0,1)$ and $(1,2)$ on the real like).
– Tobias Kildetoft
Sep 30 '15 at 13:06
1
So is there no final conclusion about their intersection?
– Kavita
Sep 30 '15 at 13:14
1
No, your conclusion that $Acap B$ is non-empty is wrong. What you can say is that there exists a point $c$ such that $c$ is a limit point in both $A$ and $B$. For example $mathbb R^+$ and $mathbb R^-$ have distance $0$, but there is no real number that is both positive and negative (but a limit point $0$ exists to both sets).
– skyking
Sep 30 '15 at 13:28
@skyking Not even that, for the limit point can be "missing" from $X$. Think of $A = {(x_1, 0) mid x_1 in mathbf R}$, and $B = {x mid x_1 x_2 = 1}$ in $X = mathbf R^2$.
– martini
Sep 30 '15 at 13:30
@martini My bad, there's a sequence in $A$ and $B$ such that $lim d(a_j, b_j) = 0$. You need $A$ and $B$ to be part of a compact set to get a limit point (unless I'm mistaken again).
– skyking
Sep 30 '15 at 13:46