Question with regards to proof about diameter of a closed ball.












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I have a question with regards to my understanding: Consider the following: If a,z $in$ X and r,s$in$ $mathbb{R}^+$ then diam(B[a,r]) $leq$ 2r. Where B[a,r] is the closed ball and X is a metric space, why is it okay to start the proof with the following: "Let x,y $in$ X and d(a,x) $leq$ r and d(a,y) $leq$ r.."? what if X or the ball has only one element? Is it because im taking arbitrary elements that means i'm not excluding the possibility that x=y? and so it won't matter if I consider even more than 2 as well? Could someone please explain further, in terms of logic (if possible?, please?)






real-analysis general-topology analysis proof-verification proof-writing







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  • 1




    $begingroup$
    You are exactly right, x = y is not excluded. And yes, this can be done for more than 2 as well.
    $endgroup$
    – Noel Lundström
    Jan 20 at 0:22












  • $begingroup$
    @NoelLundström Thank you very much, but what if the ball was empty?Is that impossible because since a metric space is non-empty, a ball cannot not be empty since it requires a metric and an element from the metric space?
    $endgroup$
    – mathsssislife
    Jan 20 at 0:26












  • $begingroup$
    This is necessary not only for the proof, but for the diameter to be well-defined at all. Otherwise, if you have a small ball in a discrete metric (which contains only its center), the diameter would be the supremum of an empty set ...
    $endgroup$
    – Henning Makholm
    Jan 20 at 0:28










  • $begingroup$
    A ball can't be empty because it must be centered on a point in the metric space.
    $endgroup$
    – Henning Makholm
    Jan 20 at 0:28










  • $begingroup$
    @HenningMakholm thank you very very much!
    $endgroup$
    – mathsssislife
    Jan 20 at 0:28
















0












$begingroup$


I have a question with regards to my understanding: Consider the following: If a,z $in$ X and r,s$in$ $mathbb{R}^+$ then diam(B[a,r]) $leq$ 2r. Where B[a,r] is the closed ball and X is a metric space, why is it okay to start the proof with the following: "Let x,y $in$ X and d(a,x) $leq$ r and d(a,y) $leq$ r.."? what if X or the ball has only one element? Is it because im taking arbitrary elements that means i'm not excluding the possibility that x=y? and so it won't matter if I consider even more than 2 as well? Could someone please explain further, in terms of logic (if possible?, please?)






real-analysis general-topology analysis proof-verification proof-writing







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You are exactly right, x = y is not excluded. And yes, this can be done for more than 2 as well.
    $endgroup$
    – Noel Lundström
    Jan 20 at 0:22












  • $begingroup$
    @NoelLundström Thank you very much, but what if the ball was empty?Is that impossible because since a metric space is non-empty, a ball cannot not be empty since it requires a metric and an element from the metric space?
    $endgroup$
    – mathsssislife
    Jan 20 at 0:26












  • $begingroup$
    This is necessary not only for the proof, but for the diameter to be well-defined at all. Otherwise, if you have a small ball in a discrete metric (which contains only its center), the diameter would be the supremum of an empty set ...
    $endgroup$
    – Henning Makholm
    Jan 20 at 0:28










  • $begingroup$
    A ball can't be empty because it must be centered on a point in the metric space.
    $endgroup$
    – Henning Makholm
    Jan 20 at 0:28










  • $begingroup$
    @HenningMakholm thank you very very much!
    $endgroup$
    – mathsssislife
    Jan 20 at 0:28














0












0








0





$begingroup$


I have a question with regards to my understanding: Consider the following: If a,z $in$ X and r,s$in$ $mathbb{R}^+$ then diam(B[a,r]) $leq$ 2r. Where B[a,r] is the closed ball and X is a metric space, why is it okay to start the proof with the following: "Let x,y $in$ X and d(a,x) $leq$ r and d(a,y) $leq$ r.."? what if X or the ball has only one element? Is it because im taking arbitrary elements that means i'm not excluding the possibility that x=y? and so it won't matter if I consider even more than 2 as well? Could someone please explain further, in terms of logic (if possible?, please?)






real-analysis general-topology analysis proof-verification proof-writing







share|cite|improve this question











$endgroup$




I have a question with regards to my understanding: Consider the following: If a,z $in$ X and r,s$in$ $mathbb{R}^+$ then diam(B[a,r]) $leq$ 2r. Where B[a,r] is the closed ball and X is a metric space, why is it okay to start the proof with the following: "Let x,y $in$ X and d(a,x) $leq$ r and d(a,y) $leq$ r.."? what if X or the ball has only one element? Is it because im taking arbitrary elements that means i'm not excluding the possibility that x=y? and so it won't matter if I consider even more than 2 as well? Could someone please explain further, in terms of logic (if possible?, please?)






real-analysis general-topology analysis proof-verification proof-writing




real-analysis general-topology analysis proof-verification proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 20 at 0:24







mathsssislife

















asked Jan 20 at 0:19









mathsssislifemathsssislife

408




408








  • 1




    $begingroup$
    You are exactly right, x = y is not excluded. And yes, this can be done for more than 2 as well.
    $endgroup$
    – Noel Lundström
    Jan 20 at 0:22












  • $begingroup$
    @NoelLundström Thank you very much, but what if the ball was empty?Is that impossible because since a metric space is non-empty, a ball cannot not be empty since it requires a metric and an element from the metric space?
    $endgroup$
    – mathsssislife
    Jan 20 at 0:26












  • $begingroup$
    This is necessary not only for the proof, but for the diameter to be well-defined at all. Otherwise, if you have a small ball in a discrete metric (which contains only its center), the diameter would be the supremum of an empty set ...
    $endgroup$
    – Henning Makholm
    Jan 20 at 0:28










  • $begingroup$
    A ball can't be empty because it must be centered on a point in the metric space.
    $endgroup$
    – Henning Makholm
    Jan 20 at 0:28










  • $begingroup$
    @HenningMakholm thank you very very much!
    $endgroup$
    – mathsssislife
    Jan 20 at 0:28














  • 1




    $begingroup$
    You are exactly right, x = y is not excluded. And yes, this can be done for more than 2 as well.
    $endgroup$
    – Noel Lundström
    Jan 20 at 0:22












  • $begingroup$
    @NoelLundström Thank you very much, but what if the ball was empty?Is that impossible because since a metric space is non-empty, a ball cannot not be empty since it requires a metric and an element from the metric space?
    $endgroup$
    – mathsssislife
    Jan 20 at 0:26












  • $begingroup$
    This is necessary not only for the proof, but for the diameter to be well-defined at all. Otherwise, if you have a small ball in a discrete metric (which contains only its center), the diameter would be the supremum of an empty set ...
    $endgroup$
    – Henning Makholm
    Jan 20 at 0:28










  • $begingroup$
    A ball can't be empty because it must be centered on a point in the metric space.
    $endgroup$
    – Henning Makholm
    Jan 20 at 0:28










  • $begingroup$
    @HenningMakholm thank you very very much!
    $endgroup$
    – mathsssislife
    Jan 20 at 0:28








1




1




$begingroup$
You are exactly right, x = y is not excluded. And yes, this can be done for more than 2 as well.
$endgroup$
– Noel Lundström
Jan 20 at 0:22






$begingroup$
You are exactly right, x = y is not excluded. And yes, this can be done for more than 2 as well.
$endgroup$
– Noel Lundström
Jan 20 at 0:22














$begingroup$
@NoelLundström Thank you very much, but what if the ball was empty?Is that impossible because since a metric space is non-empty, a ball cannot not be empty since it requires a metric and an element from the metric space?
$endgroup$
– mathsssislife
Jan 20 at 0:26






$begingroup$
@NoelLundström Thank you very much, but what if the ball was empty?Is that impossible because since a metric space is non-empty, a ball cannot not be empty since it requires a metric and an element from the metric space?
$endgroup$
– mathsssislife
Jan 20 at 0:26














$begingroup$
This is necessary not only for the proof, but for the diameter to be well-defined at all. Otherwise, if you have a small ball in a discrete metric (which contains only its center), the diameter would be the supremum of an empty set ...
$endgroup$
– Henning Makholm
Jan 20 at 0:28




$begingroup$
This is necessary not only for the proof, but for the diameter to be well-defined at all. Otherwise, if you have a small ball in a discrete metric (which contains only its center), the diameter would be the supremum of an empty set ...
$endgroup$
– Henning Makholm
Jan 20 at 0:28












$begingroup$
A ball can't be empty because it must be centered on a point in the metric space.
$endgroup$
– Henning Makholm
Jan 20 at 0:28




$begingroup$
A ball can't be empty because it must be centered on a point in the metric space.
$endgroup$
– Henning Makholm
Jan 20 at 0:28












$begingroup$
@HenningMakholm thank you very very much!
$endgroup$
– mathsssislife
Jan 20 at 0:28




$begingroup$
@HenningMakholm thank you very very much!
$endgroup$
– mathsssislife
Jan 20 at 0:28










1 Answer
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$begingroup$

You start by picking two arbitrary elements of $B[a,r]$, because we are proving the statement $forall x,y in B[a,r]: d(x,y) le 2r$ which will imply the diameter fact immediately. We make no assumptions on $x$ or $y$ beyond what is given by the definition of $B[a,r]$, namely $d(x,a) le r$. They could indeed by equal. Logically speaking, $B[a,r]$ could even be empty, as a statement of the form $forall x,y in A: phi(x)$ is always true. But of course it's not, in this particular case (but we don't need it), as $a in B[a,r]$ when $r ge 0$.



But $operatorname{diam}(A)$ is only well-defined for $A neq emptyset$ as all elements of $mathbb{R}$ are upper bounds for $emptyset={d(x,y): x,y in emptyset}$ and so the least upper bound does not exist (or is defined to be $-infty$), so it's comforting that that $B[a,r]$ is non-empty, as guaranteed by $rge 0$ (if $r <0$ the metric space axioms imply that it is empty).






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    $begingroup$

    You start by picking two arbitrary elements of $B[a,r]$, because we are proving the statement $forall x,y in B[a,r]: d(x,y) le 2r$ which will imply the diameter fact immediately. We make no assumptions on $x$ or $y$ beyond what is given by the definition of $B[a,r]$, namely $d(x,a) le r$. They could indeed by equal. Logically speaking, $B[a,r]$ could even be empty, as a statement of the form $forall x,y in A: phi(x)$ is always true. But of course it's not, in this particular case (but we don't need it), as $a in B[a,r]$ when $r ge 0$.



    But $operatorname{diam}(A)$ is only well-defined for $A neq emptyset$ as all elements of $mathbb{R}$ are upper bounds for $emptyset={d(x,y): x,y in emptyset}$ and so the least upper bound does not exist (or is defined to be $-infty$), so it's comforting that that $B[a,r]$ is non-empty, as guaranteed by $rge 0$ (if $r <0$ the metric space axioms imply that it is empty).






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      You start by picking two arbitrary elements of $B[a,r]$, because we are proving the statement $forall x,y in B[a,r]: d(x,y) le 2r$ which will imply the diameter fact immediately. We make no assumptions on $x$ or $y$ beyond what is given by the definition of $B[a,r]$, namely $d(x,a) le r$. They could indeed by equal. Logically speaking, $B[a,r]$ could even be empty, as a statement of the form $forall x,y in A: phi(x)$ is always true. But of course it's not, in this particular case (but we don't need it), as $a in B[a,r]$ when $r ge 0$.



      But $operatorname{diam}(A)$ is only well-defined for $A neq emptyset$ as all elements of $mathbb{R}$ are upper bounds for $emptyset={d(x,y): x,y in emptyset}$ and so the least upper bound does not exist (or is defined to be $-infty$), so it's comforting that that $B[a,r]$ is non-empty, as guaranteed by $rge 0$ (if $r <0$ the metric space axioms imply that it is empty).






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        You start by picking two arbitrary elements of $B[a,r]$, because we are proving the statement $forall x,y in B[a,r]: d(x,y) le 2r$ which will imply the diameter fact immediately. We make no assumptions on $x$ or $y$ beyond what is given by the definition of $B[a,r]$, namely $d(x,a) le r$. They could indeed by equal. Logically speaking, $B[a,r]$ could even be empty, as a statement of the form $forall x,y in A: phi(x)$ is always true. But of course it's not, in this particular case (but we don't need it), as $a in B[a,r]$ when $r ge 0$.



        But $operatorname{diam}(A)$ is only well-defined for $A neq emptyset$ as all elements of $mathbb{R}$ are upper bounds for $emptyset={d(x,y): x,y in emptyset}$ and so the least upper bound does not exist (or is defined to be $-infty$), so it's comforting that that $B[a,r]$ is non-empty, as guaranteed by $rge 0$ (if $r <0$ the metric space axioms imply that it is empty).






        share|cite|improve this answer









        $endgroup$



        You start by picking two arbitrary elements of $B[a,r]$, because we are proving the statement $forall x,y in B[a,r]: d(x,y) le 2r$ which will imply the diameter fact immediately. We make no assumptions on $x$ or $y$ beyond what is given by the definition of $B[a,r]$, namely $d(x,a) le r$. They could indeed by equal. Logically speaking, $B[a,r]$ could even be empty, as a statement of the form $forall x,y in A: phi(x)$ is always true. But of course it's not, in this particular case (but we don't need it), as $a in B[a,r]$ when $r ge 0$.



        But $operatorname{diam}(A)$ is only well-defined for $A neq emptyset$ as all elements of $mathbb{R}$ are upper bounds for $emptyset={d(x,y): x,y in emptyset}$ and so the least upper bound does not exist (or is defined to be $-infty$), so it's comforting that that $B[a,r]$ is non-empty, as guaranteed by $rge 0$ (if $r <0$ the metric space axioms imply that it is empty).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 20 at 7:32









        Henno BrandsmaHenno Brandsma

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