Question with regards to proof about diameter of a closed ball.
$begingroup$
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
asked Jan 20 at 0:19
mathsssislifemathsssislife
408
$endgroup$
add a comment |
$begingroup$
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
asked Jan 20 at 0:19
mathsssislifemathsssislife
408
$endgroup$
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
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@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
add a comment |
$begingroup$
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
asked Jan 20 at 0:19
mathsssislifemathsssislife
408
$endgroup$
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
asked Jan 20 at 0:19
mathsssislifemathsssislife
408
asked Jan 20 at 0:19
mathsssislifemathsssislife
408
edited Jan 20 at 0:24
mathsssislife
asked Jan 20 at 0:19
mathsssislifemathsssislife
408
asked Jan 20 at 0:19
mathsssislifemathsssislife
408
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
add a comment |
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
add a comment |
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).
answered Jan 20 at 7:32
Henno BrandsmaHenno Brandsma
110k347116
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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).
answered Jan 20 at 7:32
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
$begingroup$
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).
answered Jan 20 at 7:32
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
$begingroup$
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).
answered Jan 20 at 7:32
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
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).
answered Jan 20 at 7:32
Henno BrandsmaHenno Brandsma
110k347116
answered Jan 20 at 7:32
Henno BrandsmaHenno Brandsma
110k347116
answered Jan 20 at 7:32
Henno BrandsmaHenno Brandsma
110k347116
answered Jan 20 at 7:32
Henno BrandsmaHenno Brandsma
110k347116
110k347116
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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