Is the following statement is True/false ? reagarding closed ball
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Is the following statement is True/false ?
For $x∈mathbb{R}^n$ , let $B(x,r)$ denote the closed ball in $mathbb{R}^n$(with Euclidean norm) of radius $r$ centered at $x$. Write $B=B(0,1)$.If $f,g:B→mathbb{R}^n$ are continuous functions such that $f(x)≠g(x)$ for all $ x∈ B$, then
There exist $ϵ>0$ such that $ B(f(x), ϵ) ∩ B(g(x), ϵ)=varnothing$ for all $ x∈ B$
My attempt : i take $f(x) = x$, $g(x) = x+1$, now i visualize the diagram i take $ϵ$ as constant that greater then $0$
from the diagram i can conclude that this statement is false .
Is its True ?
any hints/solution will be appreciated
thanks u
general-topology
add a comment |
up vote
0
down vote
favorite
Is the following statement is True/false ?
For $x∈mathbb{R}^n$ , let $B(x,r)$ denote the closed ball in $mathbb{R}^n$(with Euclidean norm) of radius $r$ centered at $x$. Write $B=B(0,1)$.If $f,g:B→mathbb{R}^n$ are continuous functions such that $f(x)≠g(x)$ for all $ x∈ B$, then
There exist $ϵ>0$ such that $ B(f(x), ϵ) ∩ B(g(x), ϵ)=varnothing$ for all $ x∈ B$
My attempt : i take $f(x) = x$, $g(x) = x+1$, now i visualize the diagram i take $ϵ$ as constant that greater then $0$
from the diagram i can conclude that this statement is false .
Is its True ?
any hints/solution will be appreciated
thanks u
general-topology
1
In your example what if $epsilon = frac 14$. Then do $B(x, frac 14)$ and $B(x+1, frac 14)$ have any points in common?
– fleablood
Nov 27 at 0:06
@fleablood great logics ...i missed that
– jasmine
Nov 27 at 0:07
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is the following statement is True/false ?
For $x∈mathbb{R}^n$ , let $B(x,r)$ denote the closed ball in $mathbb{R}^n$(with Euclidean norm) of radius $r$ centered at $x$. Write $B=B(0,1)$.If $f,g:B→mathbb{R}^n$ are continuous functions such that $f(x)≠g(x)$ for all $ x∈ B$, then
There exist $ϵ>0$ such that $ B(f(x), ϵ) ∩ B(g(x), ϵ)=varnothing$ for all $ x∈ B$
My attempt : i take $f(x) = x$, $g(x) = x+1$, now i visualize the diagram i take $ϵ$ as constant that greater then $0$
from the diagram i can conclude that this statement is false .
Is its True ?
any hints/solution will be appreciated
thanks u
general-topology
Is the following statement is True/false ?
For $x∈mathbb{R}^n$ , let $B(x,r)$ denote the closed ball in $mathbb{R}^n$(with Euclidean norm) of radius $r$ centered at $x$. Write $B=B(0,1)$.If $f,g:B→mathbb{R}^n$ are continuous functions such that $f(x)≠g(x)$ for all $ x∈ B$, then
There exist $ϵ>0$ such that $ B(f(x), ϵ) ∩ B(g(x), ϵ)=varnothing$ for all $ x∈ B$
My attempt : i take $f(x) = x$, $g(x) = x+1$, now i visualize the diagram i take $ϵ$ as constant that greater then $0$
from the diagram i can conclude that this statement is false .
Is its True ?
any hints/solution will be appreciated
thanks u
general-topology
general-topology
asked Nov 26 at 20:54
jasmine
1,403416
1,403416
1
In your example what if $epsilon = frac 14$. Then do $B(x, frac 14)$ and $B(x+1, frac 14)$ have any points in common?
– fleablood
Nov 27 at 0:06
@fleablood great logics ...i missed that
– jasmine
Nov 27 at 0:07
add a comment |
1
In your example what if $epsilon = frac 14$. Then do $B(x, frac 14)$ and $B(x+1, frac 14)$ have any points in common?
– fleablood
Nov 27 at 0:06
@fleablood great logics ...i missed that
– jasmine
Nov 27 at 0:07
1
1
In your example what if $epsilon = frac 14$. Then do $B(x, frac 14)$ and $B(x+1, frac 14)$ have any points in common?
– fleablood
Nov 27 at 0:06
In your example what if $epsilon = frac 14$. Then do $B(x, frac 14)$ and $B(x+1, frac 14)$ have any points in common?
– fleablood
Nov 27 at 0:06
@fleablood great logics ...i missed that
– jasmine
Nov 27 at 0:07
@fleablood great logics ...i missed that
– jasmine
Nov 27 at 0:07
add a comment |
2 Answers
2
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up vote
1
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accepted
Your argument is not correct. If $0<epsilon <1$ there cannot be any point common to the two balls. To give a counterexample take $f(x)=(1-|x|,0,..,0)$ and $g(x)=(0,0,..,0)$ for all $x in B$. Then $f(x) neq g(x)$ for all $x in B$ and there is no $epsilon$ such that $B(f(x),epsilon)cap B(g(x),epsilon)=emptyset$ for all $x in B$. To see this take $x =(1-epsilon /2,0,0...,0)$ and note that $(0,cdots,0)$ is a point in the intersection of the two balls.
add a comment |
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$f(x) ne g(x)$ for any $xin B$. So $d(f(x), g(x)) > 0$ for all $x in B$. So $A= {d(f(x),g(x))|xin B}$ is non empty and bounded below by $0$ so $inf A$ exist.
If $inf A > 0$ then for any $0< epsilon < frac 12inf A$. We'd be done. The statement is true[1].
But what if $inf A = 0$. Then the statement is false[2].
So we can surely find some acceptable $f$ and $g$ where $inf {d(f(x),g(x))|xin B} = 0$.[3]
=====
1)
If $c in B(f(x),epsilon)cap B(g(x),epsilon)$ then by triangle inequality $d(f(x),g(x)) le d(f(x),c) + d(g(c),c) < epsilon + epsilon < d(f(x),g(x))$ that 's a contradiction.
2)
For any $epsilon > 0$ there exists an $x$ where $d(f(x), g(x)) < epsilon$ so $g(x) in B(f(x), epsilon)$ and $f(x) in B(g(x),epsilon)$ and $g(x) in B(g(x),epsilon)$ and $f(x) in B(f(x), epsilon)$ so $f(x), g(x) in B(g(x),epsilon)cap B(f(x), epsilon)$.
3)
Let $f(x)=x$ be the identity function let $g(x)= (2,0,0,0,0,....) - x$. Let for any $1 > delta > 0$ then $z= (1-frac 12delta, 0,0,0,.....) in B$ and $d(f(z), g(z)) = d((1-frac 12delta,0,0,0....) ,(1+frac 12delta,0,0,0,....)) = delta$.
So for all $delta > 0$ there is a $z in B$ so that $d(f(x), g(x)) = delta$ so $inf {d(f(x),g(x))|xin B} = 0$.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your argument is not correct. If $0<epsilon <1$ there cannot be any point common to the two balls. To give a counterexample take $f(x)=(1-|x|,0,..,0)$ and $g(x)=(0,0,..,0)$ for all $x in B$. Then $f(x) neq g(x)$ for all $x in B$ and there is no $epsilon$ such that $B(f(x),epsilon)cap B(g(x),epsilon)=emptyset$ for all $x in B$. To see this take $x =(1-epsilon /2,0,0...,0)$ and note that $(0,cdots,0)$ is a point in the intersection of the two balls.
add a comment |
up vote
1
down vote
accepted
Your argument is not correct. If $0<epsilon <1$ there cannot be any point common to the two balls. To give a counterexample take $f(x)=(1-|x|,0,..,0)$ and $g(x)=(0,0,..,0)$ for all $x in B$. Then $f(x) neq g(x)$ for all $x in B$ and there is no $epsilon$ such that $B(f(x),epsilon)cap B(g(x),epsilon)=emptyset$ for all $x in B$. To see this take $x =(1-epsilon /2,0,0...,0)$ and note that $(0,cdots,0)$ is a point in the intersection of the two balls.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your argument is not correct. If $0<epsilon <1$ there cannot be any point common to the two balls. To give a counterexample take $f(x)=(1-|x|,0,..,0)$ and $g(x)=(0,0,..,0)$ for all $x in B$. Then $f(x) neq g(x)$ for all $x in B$ and there is no $epsilon$ such that $B(f(x),epsilon)cap B(g(x),epsilon)=emptyset$ for all $x in B$. To see this take $x =(1-epsilon /2,0,0...,0)$ and note that $(0,cdots,0)$ is a point in the intersection of the two balls.
Your argument is not correct. If $0<epsilon <1$ there cannot be any point common to the two balls. To give a counterexample take $f(x)=(1-|x|,0,..,0)$ and $g(x)=(0,0,..,0)$ for all $x in B$. Then $f(x) neq g(x)$ for all $x in B$ and there is no $epsilon$ such that $B(f(x),epsilon)cap B(g(x),epsilon)=emptyset$ for all $x in B$. To see this take $x =(1-epsilon /2,0,0...,0)$ and note that $(0,cdots,0)$ is a point in the intersection of the two balls.
answered Nov 26 at 23:59
Kavi Rama Murthy
46k31854
46k31854
add a comment |
add a comment |
up vote
1
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$f(x) ne g(x)$ for any $xin B$. So $d(f(x), g(x)) > 0$ for all $x in B$. So $A= {d(f(x),g(x))|xin B}$ is non empty and bounded below by $0$ so $inf A$ exist.
If $inf A > 0$ then for any $0< epsilon < frac 12inf A$. We'd be done. The statement is true[1].
But what if $inf A = 0$. Then the statement is false[2].
So we can surely find some acceptable $f$ and $g$ where $inf {d(f(x),g(x))|xin B} = 0$.[3]
=====
1)
If $c in B(f(x),epsilon)cap B(g(x),epsilon)$ then by triangle inequality $d(f(x),g(x)) le d(f(x),c) + d(g(c),c) < epsilon + epsilon < d(f(x),g(x))$ that 's a contradiction.
2)
For any $epsilon > 0$ there exists an $x$ where $d(f(x), g(x)) < epsilon$ so $g(x) in B(f(x), epsilon)$ and $f(x) in B(g(x),epsilon)$ and $g(x) in B(g(x),epsilon)$ and $f(x) in B(f(x), epsilon)$ so $f(x), g(x) in B(g(x),epsilon)cap B(f(x), epsilon)$.
3)
Let $f(x)=x$ be the identity function let $g(x)= (2,0,0,0,0,....) - x$. Let for any $1 > delta > 0$ then $z= (1-frac 12delta, 0,0,0,.....) in B$ and $d(f(z), g(z)) = d((1-frac 12delta,0,0,0....) ,(1+frac 12delta,0,0,0,....)) = delta$.
So for all $delta > 0$ there is a $z in B$ so that $d(f(x), g(x)) = delta$ so $inf {d(f(x),g(x))|xin B} = 0$.
add a comment |
up vote
1
down vote
$f(x) ne g(x)$ for any $xin B$. So $d(f(x), g(x)) > 0$ for all $x in B$. So $A= {d(f(x),g(x))|xin B}$ is non empty and bounded below by $0$ so $inf A$ exist.
If $inf A > 0$ then for any $0< epsilon < frac 12inf A$. We'd be done. The statement is true[1].
But what if $inf A = 0$. Then the statement is false[2].
So we can surely find some acceptable $f$ and $g$ where $inf {d(f(x),g(x))|xin B} = 0$.[3]
=====
1)
If $c in B(f(x),epsilon)cap B(g(x),epsilon)$ then by triangle inequality $d(f(x),g(x)) le d(f(x),c) + d(g(c),c) < epsilon + epsilon < d(f(x),g(x))$ that 's a contradiction.
2)
For any $epsilon > 0$ there exists an $x$ where $d(f(x), g(x)) < epsilon$ so $g(x) in B(f(x), epsilon)$ and $f(x) in B(g(x),epsilon)$ and $g(x) in B(g(x),epsilon)$ and $f(x) in B(f(x), epsilon)$ so $f(x), g(x) in B(g(x),epsilon)cap B(f(x), epsilon)$.
3)
Let $f(x)=x$ be the identity function let $g(x)= (2,0,0,0,0,....) - x$. Let for any $1 > delta > 0$ then $z= (1-frac 12delta, 0,0,0,.....) in B$ and $d(f(z), g(z)) = d((1-frac 12delta,0,0,0....) ,(1+frac 12delta,0,0,0,....)) = delta$.
So for all $delta > 0$ there is a $z in B$ so that $d(f(x), g(x)) = delta$ so $inf {d(f(x),g(x))|xin B} = 0$.
add a comment |
up vote
1
down vote
up vote
1
down vote
$f(x) ne g(x)$ for any $xin B$. So $d(f(x), g(x)) > 0$ for all $x in B$. So $A= {d(f(x),g(x))|xin B}$ is non empty and bounded below by $0$ so $inf A$ exist.
If $inf A > 0$ then for any $0< epsilon < frac 12inf A$. We'd be done. The statement is true[1].
But what if $inf A = 0$. Then the statement is false[2].
So we can surely find some acceptable $f$ and $g$ where $inf {d(f(x),g(x))|xin B} = 0$.[3]
=====
1)
If $c in B(f(x),epsilon)cap B(g(x),epsilon)$ then by triangle inequality $d(f(x),g(x)) le d(f(x),c) + d(g(c),c) < epsilon + epsilon < d(f(x),g(x))$ that 's a contradiction.
2)
For any $epsilon > 0$ there exists an $x$ where $d(f(x), g(x)) < epsilon$ so $g(x) in B(f(x), epsilon)$ and $f(x) in B(g(x),epsilon)$ and $g(x) in B(g(x),epsilon)$ and $f(x) in B(f(x), epsilon)$ so $f(x), g(x) in B(g(x),epsilon)cap B(f(x), epsilon)$.
3)
Let $f(x)=x$ be the identity function let $g(x)= (2,0,0,0,0,....) - x$. Let for any $1 > delta > 0$ then $z= (1-frac 12delta, 0,0,0,.....) in B$ and $d(f(z), g(z)) = d((1-frac 12delta,0,0,0....) ,(1+frac 12delta,0,0,0,....)) = delta$.
So for all $delta > 0$ there is a $z in B$ so that $d(f(x), g(x)) = delta$ so $inf {d(f(x),g(x))|xin B} = 0$.
$f(x) ne g(x)$ for any $xin B$. So $d(f(x), g(x)) > 0$ for all $x in B$. So $A= {d(f(x),g(x))|xin B}$ is non empty and bounded below by $0$ so $inf A$ exist.
If $inf A > 0$ then for any $0< epsilon < frac 12inf A$. We'd be done. The statement is true[1].
But what if $inf A = 0$. Then the statement is false[2].
So we can surely find some acceptable $f$ and $g$ where $inf {d(f(x),g(x))|xin B} = 0$.[3]
=====
1)
If $c in B(f(x),epsilon)cap B(g(x),epsilon)$ then by triangle inequality $d(f(x),g(x)) le d(f(x),c) + d(g(c),c) < epsilon + epsilon < d(f(x),g(x))$ that 's a contradiction.
2)
For any $epsilon > 0$ there exists an $x$ where $d(f(x), g(x)) < epsilon$ so $g(x) in B(f(x), epsilon)$ and $f(x) in B(g(x),epsilon)$ and $g(x) in B(g(x),epsilon)$ and $f(x) in B(f(x), epsilon)$ so $f(x), g(x) in B(g(x),epsilon)cap B(f(x), epsilon)$.
3)
Let $f(x)=x$ be the identity function let $g(x)= (2,0,0,0,0,....) - x$. Let for any $1 > delta > 0$ then $z= (1-frac 12delta, 0,0,0,.....) in B$ and $d(f(z), g(z)) = d((1-frac 12delta,0,0,0....) ,(1+frac 12delta,0,0,0,....)) = delta$.
So for all $delta > 0$ there is a $z in B$ so that $d(f(x), g(x)) = delta$ so $inf {d(f(x),g(x))|xin B} = 0$.
answered Nov 27 at 0:46
fleablood
67.3k22684
67.3k22684
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1
In your example what if $epsilon = frac 14$. Then do $B(x, frac 14)$ and $B(x+1, frac 14)$ have any points in common?
– fleablood
Nov 27 at 0:06
@fleablood great logics ...i missed that
– jasmine
Nov 27 at 0:07