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[SOLVED] metric space

dwsmith

Well-known member
Feb 1, 2012
1,673
Consider the metric spaces in $\mathbb{R}^n$:
$$
d_1(\mathbf{x},\mathbf{y}) = \max\limits_{1\leq i\leq n}|x_i - y_i|
$$
Prove that the ball $B(\mathbf{a},r)$ has the geometric appearance indicated:


In $(\mathbb{R}^2,d_1)$, a square with sides parallel to the coordinate axes.


Consider the ball $B((x_1,y_1),r)$ where $r > 0$.
Let $(x_2,y_2)\in B((x_1,y_1),r)$.
We know that $|x_1 - x_2| < r\iff -r < x_1 - x_2 < r$ and $|y_1 - y_2| < r\iff -r < y_1 - y_2 < r$ by construction and which forms a square with sides parallel to the coordinate axes.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
So I am looking at an example I created.

$(1,2)\in B((2,2),3)$

Then $d = |1| < 3$. So does that mean go 2 units above/below the center and 2 units right/left of the center?

Code:
   ------x------
  :               :
  x      x      x
  :               :
  -------x------
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Consider the metric spaces in $\mathbb{R}^n$:
$$
d_1(\mathbf{x},\mathbf{y}) = \max\limits_{1\leq i\leq n}|x_i - y_i|
$$
Prove that the ball $B(\mathbf{a},r)$ has the geometric appearance indicated:


In $(\mathbb{R}^2,d_1)$, a square with sides parallel to the coordinate axes.


Consider the ball $B((x_1,y_1),r)$ where $r > 0$.
Let $(x_2,y_2)\in B((x_1,y_1),r)$.
We know that $|x_1 - x_2| < r\iff -r < x_1 - x_2 < r$ and $|y_1 - y_2| < r\iff -r < y_1 - y_2 < r$ by construction and which forms a square with sides parallel to the coordinate axes.
This is wrong since it has to be $ |x_1 - y_1|$.
I don't understand how to do this.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Consider the case $B((0,0),1)$ where $(x_1,x_2)\in B((0,0),1)$.
Therefore, $|x_1|<1$ and $|x_2|<1$.
Why does that mean we have square? Aren't x_1 and x_2 on the x axis?
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
Consider the case $B((0,0),1)$ where $(x_1,x_2)\in B((0,0),1)$.
Therefore, $|x_1|<1$ and $|x_2|<1$.
Why does that mean we have square? Aren't x_1 and x_2 on the x axis?
$x_1$ is not a point, it's the horizontal coordinate of a point. The inequalities $|x_1|<1$ and $|x_2|<1$ mean that points are allowed to have horizontal coordinates between -1 and 1 and (independent of the horizontal coordinate) vertical coordinates between -1 and 1. These points form a square with side 2.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
$x_1$ is not a point, it's the horizontal coordinate of a point. The inequalities $|x_1|<1$ and $|x_2|<1$ mean that points are allowed to have horizontal coordinates between -1 and 1 and (independent of the horizontal coordinate) vertical coordinates between -1 and 1. These points form a square with side 2.
I understand the horizontal but where do the vertical lines come from?
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
Consider the metric spaces in $\mathbb{R}^n$:
$$
d_1(\mathbf{x},\mathbf{y}) = \max\limits_{1\leq i\leq n}|x_i - y_i|
$$
Prove that the ball $B(\mathbf{a},r)$ has the geometric appearance indicated:


In $(\mathbb{R}^2,d_1)$, a square with sides parallel to the coordinate axes.
In this problem, $\mathbf{x}=(x_1,\dots,x_n)$ represents a point in $\mathbb{R}^n$. When n = 2, the point is $(x_1, x_2)$. This can be represented as a point on a 2D coordinate plain by making $x_1$ the horizontal coordinate and $x_2$ the vertical coordinate. A point $\mathbf{y}=(y_1,y_2)$ can also be represented on the plain by making $y_1$ the horizontal coordinate and $y_2$ the vertical coordinate.