$A$ is compact and closed then prove …
$begingroup$
I am taking an introductory real analysis course and I have difficulty understanding and solving the problem below .Is it trying to say that we have an infimum for the distance between every two points in $A$ and another arbitrary set?
$A$ is compact and closed then prove there exist an $a$ member of $A$ and $b$ a member of $B$ such that $d(a,b) = d(x,y)$ ($x$ is a member of $A$ and $y$ a member of $B$).
real-analysis compactness closed-form
edited Dec 10 '18 at 6:32
Ali
1,9682520
asked Dec 10 '18 at 6:04
PegiPegi
32
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add a comment |
$begingroup$
I am taking an introductory real analysis course and I have difficulty understanding and solving the problem below .Is it trying to say that we have an infimum for the distance between every two points in $A$ and another arbitrary set?
$A$ is compact and closed then prove there exist an $a$ member of $A$ and $b$ a member of $B$ such that $d(a,b) = d(x,y)$ ($x$ is a member of $A$ and $y$ a member of $B$).
real-analysis compactness closed-form
edited Dec 10 '18 at 6:32
Ali
1,9682520
asked Dec 10 '18 at 6:04
PegiPegi
32
$endgroup$
add a comment |
$begingroup$
I am taking an introductory real analysis course and I have difficulty understanding and solving the problem below .Is it trying to say that we have an infimum for the distance between every two points in $A$ and another arbitrary set?
$A$ is compact and closed then prove there exist an $a$ member of $A$ and $b$ a member of $B$ such that $d(a,b) = d(x,y)$ ($x$ is a member of $A$ and $y$ a member of $B$).
real-analysis compactness closed-form
edited Dec 10 '18 at 6:32
Ali
1,9682520
asked Dec 10 '18 at 6:04
PegiPegi
32
$endgroup$
I am taking an introductory real analysis course and I have difficulty understanding and solving the problem below .Is it trying to say that we have an infimum for the distance between every two points in $A$ and another arbitrary set?
$A$ is compact and closed then prove there exist an $a$ member of $A$ and $b$ a member of $B$ such that $d(a,b) = d(x,y)$ ($x$ is a member of $A$ and $y$ a member of $B$).
real-analysis compactness closed-form
real-analysis compactness closed-form
edited Dec 10 '18 at 6:32
Ali
1,9682520
asked Dec 10 '18 at 6:04
PegiPegi
32
edited Dec 10 '18 at 6:32
Ali
1,9682520
asked Dec 10 '18 at 6:04
PegiPegi
32
edited Dec 10 '18 at 6:32
Ali
1,9682520
edited Dec 10 '18 at 6:32
Ali
1,9682520
edited Dec 10 '18 at 6:32
Ali
1,9682520
1,9682520
asked Dec 10 '18 at 6:04
PegiPegi
32
asked Dec 10 '18 at 6:04
PegiPegi
32
asked Dec 10 '18 at 6:04
PegiPegi
32
32
add a comment |
add a comment |
1 Answer
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For $A,B$ two non-empty sets, $d(A,B):= inf {d(x,y): x in A, y in B}$ always exists, as we have a set of real numbers that is non-empty and bounded below (by $0$ trivially). The same holds for $d(x,B) = inf {d(x,b); b in B}$ as a special case.
However, we can note that the function $f_B: x to d(x,B)$ is continuous for $x in X$ (as $|d(x,B) - d(x', B)| le d(x,x')$ by the triangle inequality, so the function is contractive hence uniformly continuous) and so $f_B$ assumes a minimum and maximum value on the compact subset $A$.
If we do not assume anything on $B$ we need not always have such $a,b$ as asked for: $A = [0,1], B = (1,2)$ is an example where $d(A,B)=0$ and $A$ is compact. If $B$ too is compact we do have such $a,b$, e.g. In many cases $B$ being closed is also sufficient besides $A$ compact.
answered Dec 10 '18 at 6:47
Henno BrandsmaHenno Brandsma
108k347114
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$begingroup$
For $A,B$ two non-empty sets, $d(A,B):= inf {d(x,y): x in A, y in B}$ always exists, as we have a set of real numbers that is non-empty and bounded below (by $0$ trivially). The same holds for $d(x,B) = inf {d(x,b); b in B}$ as a special case.
However, we can note that the function $f_B: x to d(x,B)$ is continuous for $x in X$ (as $|d(x,B) - d(x', B)| le d(x,x')$ by the triangle inequality, so the function is contractive hence uniformly continuous) and so $f_B$ assumes a minimum and maximum value on the compact subset $A$.
If we do not assume anything on $B$ we need not always have such $a,b$ as asked for: $A = [0,1], B = (1,2)$ is an example where $d(A,B)=0$ and $A$ is compact. If $B$ too is compact we do have such $a,b$, e.g. In many cases $B$ being closed is also sufficient besides $A$ compact.
answered Dec 10 '18 at 6:47
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
add a comment |
$begingroup$
For $A,B$ two non-empty sets, $d(A,B):= inf {d(x,y): x in A, y in B}$ always exists, as we have a set of real numbers that is non-empty and bounded below (by $0$ trivially). The same holds for $d(x,B) = inf {d(x,b); b in B}$ as a special case.
However, we can note that the function $f_B: x to d(x,B)$ is continuous for $x in X$ (as $|d(x,B) - d(x', B)| le d(x,x')$ by the triangle inequality, so the function is contractive hence uniformly continuous) and so $f_B$ assumes a minimum and maximum value on the compact subset $A$.
If we do not assume anything on $B$ we need not always have such $a,b$ as asked for: $A = [0,1], B = (1,2)$ is an example where $d(A,B)=0$ and $A$ is compact. If $B$ too is compact we do have such $a,b$, e.g. In many cases $B$ being closed is also sufficient besides $A$ compact.
answered Dec 10 '18 at 6:47
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
add a comment |
$begingroup$
For $A,B$ two non-empty sets, $d(A,B):= inf {d(x,y): x in A, y in B}$ always exists, as we have a set of real numbers that is non-empty and bounded below (by $0$ trivially). The same holds for $d(x,B) = inf {d(x,b); b in B}$ as a special case.
However, we can note that the function $f_B: x to d(x,B)$ is continuous for $x in X$ (as $|d(x,B) - d(x', B)| le d(x,x')$ by the triangle inequality, so the function is contractive hence uniformly continuous) and so $f_B$ assumes a minimum and maximum value on the compact subset $A$.
If we do not assume anything on $B$ we need not always have such $a,b$ as asked for: $A = [0,1], B = (1,2)$ is an example where $d(A,B)=0$ and $A$ is compact. If $B$ too is compact we do have such $a,b$, e.g. In many cases $B$ being closed is also sufficient besides $A$ compact.
answered Dec 10 '18 at 6:47
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
For $A,B$ two non-empty sets, $d(A,B):= inf {d(x,y): x in A, y in B}$ always exists, as we have a set of real numbers that is non-empty and bounded below (by $0$ trivially). The same holds for $d(x,B) = inf {d(x,b); b in B}$ as a special case.
However, we can note that the function $f_B: x to d(x,B)$ is continuous for $x in X$ (as $|d(x,B) - d(x', B)| le d(x,x')$ by the triangle inequality, so the function is contractive hence uniformly continuous) and so $f_B$ assumes a minimum and maximum value on the compact subset $A$.
If we do not assume anything on $B$ we need not always have such $a,b$ as asked for: $A = [0,1], B = (1,2)$ is an example where $d(A,B)=0$ and $A$ is compact. If $B$ too is compact we do have such $a,b$, e.g. In many cases $B$ being closed is also sufficient besides $A$ compact.
answered Dec 10 '18 at 6:47
Henno BrandsmaHenno Brandsma
108k347114
answered Dec 10 '18 at 6:47
Henno BrandsmaHenno Brandsma
108k347114
answered Dec 10 '18 at 6:47
Henno BrandsmaHenno Brandsma
108k347114
answered Dec 10 '18 at 6:47
Henno BrandsmaHenno Brandsma
108k347114
108k347114
add a comment |
add a comment |
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