Intersection of balls in $l^{infty}$












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Must it be the case that the intersection of any two closed balls in $l^{infty}$ either is empty, a single point, or contains a non-empty open ball?






real-analysis general-topology functional-analysis







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  • 5




    $begingroup$
    Hint: consider the closed balls of radius 1 centered at the points $(0,0,0,0,dots)$ and $(2,0,0,0,dots)$.
    $endgroup$
    – Nate Eldredge
    Dec 24 '18 at 1:07
















1












$begingroup$


Must it be the case that the intersection of any two closed balls in $l^{infty}$ either is empty, a single point, or contains a non-empty open ball?






real-analysis general-topology functional-analysis







share|cite|improve this question









$endgroup$








  • 5




    $begingroup$
    Hint: consider the closed balls of radius 1 centered at the points $(0,0,0,0,dots)$ and $(2,0,0,0,dots)$.
    $endgroup$
    – Nate Eldredge
    Dec 24 '18 at 1:07














1












1








1


1



$begingroup$


Must it be the case that the intersection of any two closed balls in $l^{infty}$ either is empty, a single point, or contains a non-empty open ball?






real-analysis general-topology functional-analysis







share|cite|improve this question









$endgroup$




Must it be the case that the intersection of any two closed balls in $l^{infty}$ either is empty, a single point, or contains a non-empty open ball?






real-analysis general-topology functional-analysis




real-analysis general-topology functional-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 24 '18 at 0:04









Andrew LizarragaAndrew Lizarraga

4215




4215








  • 5




    $begingroup$
    Hint: consider the closed balls of radius 1 centered at the points $(0,0,0,0,dots)$ and $(2,0,0,0,dots)$.
    $endgroup$
    – Nate Eldredge
    Dec 24 '18 at 1:07














  • 5




    $begingroup$
    Hint: consider the closed balls of radius 1 centered at the points $(0,0,0,0,dots)$ and $(2,0,0,0,dots)$.
    $endgroup$
    – Nate Eldredge
    Dec 24 '18 at 1:07








5




5




$begingroup$
Hint: consider the closed balls of radius 1 centered at the points $(0,0,0,0,dots)$ and $(2,0,0,0,dots)$.
$endgroup$
– Nate Eldredge
Dec 24 '18 at 1:07




$begingroup$
Hint: consider the closed balls of radius 1 centered at the points $(0,0,0,0,dots)$ and $(2,0,0,0,dots)$.
$endgroup$
– Nate Eldredge
Dec 24 '18 at 1:07










1 Answer
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$begingroup$

To expand Nate's comment into a full answer: let us set $x_1:=(0,0,0,ldots )$ and $x_2:=(2,0,0,0,ldots )$ and then $$ bar{B_i}(x_i,1):={ yin ell_infty : |y-x|_infty leq 1 }$$
Since $|x|_infty = max_{iin {mathbb N}} |x_i|$ we see that the point $(1,0,0,0,ldots )$ will lie in both $bar{B_1}$ and $bar{B_2}$. In fact, you should be easily able to show that
$$bar{B_1} cap bar{B_2} = { (1, y_1, y_2, y_3, ldots ) : |y_i|leq 1 : forall i in {mathbb N} } $$
So: the intersection is




  1. clearly non-empty

  2. contains infinitely many points

  3. does not contain an open ball (because the first co-ordinate of all the points in the intersection is forced to be $1$.






share|cite|improve this answer









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    1












    $begingroup$

    To expand Nate's comment into a full answer: let us set $x_1:=(0,0,0,ldots )$ and $x_2:=(2,0,0,0,ldots )$ and then $$ bar{B_i}(x_i,1):={ yin ell_infty : |y-x|_infty leq 1 }$$
    Since $|x|_infty = max_{iin {mathbb N}} |x_i|$ we see that the point $(1,0,0,0,ldots )$ will lie in both $bar{B_1}$ and $bar{B_2}$. In fact, you should be easily able to show that
    $$bar{B_1} cap bar{B_2} = { (1, y_1, y_2, y_3, ldots ) : |y_i|leq 1 : forall i in {mathbb N} } $$
    So: the intersection is




    1. clearly non-empty

    2. contains infinitely many points

    3. does not contain an open ball (because the first co-ordinate of all the points in the intersection is forced to be $1$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      To expand Nate's comment into a full answer: let us set $x_1:=(0,0,0,ldots )$ and $x_2:=(2,0,0,0,ldots )$ and then $$ bar{B_i}(x_i,1):={ yin ell_infty : |y-x|_infty leq 1 }$$
      Since $|x|_infty = max_{iin {mathbb N}} |x_i|$ we see that the point $(1,0,0,0,ldots )$ will lie in both $bar{B_1}$ and $bar{B_2}$. In fact, you should be easily able to show that
      $$bar{B_1} cap bar{B_2} = { (1, y_1, y_2, y_3, ldots ) : |y_i|leq 1 : forall i in {mathbb N} } $$
      So: the intersection is




      1. clearly non-empty

      2. contains infinitely many points

      3. does not contain an open ball (because the first co-ordinate of all the points in the intersection is forced to be $1$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        To expand Nate's comment into a full answer: let us set $x_1:=(0,0,0,ldots )$ and $x_2:=(2,0,0,0,ldots )$ and then $$ bar{B_i}(x_i,1):={ yin ell_infty : |y-x|_infty leq 1 }$$
        Since $|x|_infty = max_{iin {mathbb N}} |x_i|$ we see that the point $(1,0,0,0,ldots )$ will lie in both $bar{B_1}$ and $bar{B_2}$. In fact, you should be easily able to show that
        $$bar{B_1} cap bar{B_2} = { (1, y_1, y_2, y_3, ldots ) : |y_i|leq 1 : forall i in {mathbb N} } $$
        So: the intersection is




        1. clearly non-empty

        2. contains infinitely many points

        3. does not contain an open ball (because the first co-ordinate of all the points in the intersection is forced to be $1$.






        share|cite|improve this answer









        $endgroup$



        To expand Nate's comment into a full answer: let us set $x_1:=(0,0,0,ldots )$ and $x_2:=(2,0,0,0,ldots )$ and then $$ bar{B_i}(x_i,1):={ yin ell_infty : |y-x|_infty leq 1 }$$
        Since $|x|_infty = max_{iin {mathbb N}} |x_i|$ we see that the point $(1,0,0,0,ldots )$ will lie in both $bar{B_1}$ and $bar{B_2}$. In fact, you should be easily able to show that
        $$bar{B_1} cap bar{B_2} = { (1, y_1, y_2, y_3, ldots ) : |y_i|leq 1 : forall i in {mathbb N} } $$
        So: the intersection is




        1. clearly non-empty

        2. contains infinitely many points

        3. does not contain an open ball (because the first co-ordinate of all the points in the intersection is forced to be $1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 24 '18 at 15:37









        postmortespostmortes

        2,04121119




        2,04121119






























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