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 $0$enter image description here



from the diagram i can conclude that this statement is false .



Is its True ?



any hints/solution will be appreciated



thanks u










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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















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 $0$enter image description here



from the diagram i can conclude that this statement is false .



Is its True ?



any hints/solution will be appreciated



thanks u










share|cite|improve this question


















  • 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













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 $0$enter image description here



from the diagram i can conclude that this statement is false .



Is its True ?



any hints/solution will be appreciated



thanks u










share|cite|improve this question













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 $0$enter image description here



from the diagram i can conclude that this statement is false .



Is its True ?



any hints/solution will be appreciated



thanks u







general-topology






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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














  • 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










2 Answers
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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.






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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 $inf {d(f(x),g(x))|xin B} = 0$.







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      2 Answers
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      2 Answers
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      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.






      share|cite|improve this answer

























        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.






        share|cite|improve this answer























          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.






          share|cite|improve this answer












          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.







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          share|cite|improve this answer










          answered Nov 26 at 23:59









          Kavi Rama Murthy

          46k31854




          46k31854






















              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 $inf {d(f(x),g(x))|xin B} = 0$.







              share|cite|improve this answer

























                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 $inf {d(f(x),g(x))|xin B} = 0$.







                share|cite|improve this answer























                  up vote
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                  down vote










                  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 $inf {d(f(x),g(x))|xin B} = 0$.







                  share|cite|improve this answer












                  $f(x) ne g(x)$ 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 $inf {d(f(x),g(x))|xin B} = 0$.








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                  answered Nov 27 at 0:46









                  fleablood

                  67.3k22684




                  67.3k22684






























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