# metric space: convergent sequence

• Feb 10th 2009, 11:23 AM
pkr
metric space: convergent sequence
X = R^2

consider two distances defined as:

d1((x1,y1),(x2,y2)) = sqrt[(x1-x2)² + (y1-y2)²]

d2((x1,y1),(x2,y2)) = |x1-x2| + |y1-y2|

Show that a sequence is convergent for the distance d1 if and only if it is convergent for the distance d2.
• Feb 11th 2009, 12:40 PM
HallsofIvy
Quote:

Originally Posted by pkr
X = R^2

consider two distances defined as:

d1((x1,y1),(x2,y2)) = sqrt[(x1-x2)² + (y1-y2)²]

d2((x1,y1),(x2,y2)) = |x1-x2| + |y1-y2|

Show that a sequence is convergent for the distance d1 if and only if it is convergent for the distance d2.

Look at the "neighborhoods" defined by these. Take (x2,y2)= 0. All (x,y) such that d1((x,y),(0, 0))= 1 satisfy $\sqrt{x^2+ y^2}= 1$ which is equivalent to $x^2+ y^2= 1$. That set is a circle of radius 1, center 0. The set of all p such that d1(p,q)< r, the "r neighborhood of q", for fixed point q is the interior of the circle with center q, radius r.

d2 is a little harder. All (x,y) such that d((x,y),(0,0))= 1 satisfy |x|+ |y|= 1. To solve that, break it into 4 cases:
a) $x\ge 0$, [tex]y\ge 0[/itex]
In this case the equation is x+ y= 1 which is a straight line through (1, 0) and (0, 1). Since x and y must be non-negative, it is the closed line segment with endpoints (1, 0), (0, 1), in the first quadrant.
b) [tex]x\ge 0[tex], y< 0
In this case the equation is x- y= 1 which is a straight line through (1, 0) and (0, -1). Since x must be non-negative and y must be negative, it is the line segment with endpoints (1, 0), (0, -1), in the fourth quadrant.
c) x< 0, $y\ge 0$
In this case the equation is -x+ y= 1 which is a straight line through (-1, 0) and (0, 1). Since x must be negative and y must be non-negative, it is the line segment with endpoints (-1, 0), (0, 1) in the second quadrant.
d) x< 0, y< 0
In this case the equation is -x- y= 1 which is a straight line through (-1, 0) and (0, -1). Since x and y both must be negative, it is the line segment with endpoints (-1, 0), (0, -1) in the fourth quadrant.

Putting all those together we have a square with vertices at (1,0), (0, 1), (0, -1) and (-1, 0) (diagonals parallel to the x and y axes) (a "diamond").

The set of all p such that d2(p,q)< r, the "r neighborhood of q", for fixed point q is the interior of the square "diamond" on q with diagonal length 2r.

What you want to do is this: show that, given any disc centered on q, of radius r, there exist some smaller "diamond" centered on q that is completely inside the disk and also that, given any "diamond" centered on q with diagonal length there exist some smaller disk centered on q completely inside the "diamond".

That way, in any proof of "limit" properties you can replace either type of neighborhood with the other.

By the way, there exist a third metric, d3((x1,y1),(x2,y2))= max(|x1-x2|, |y1- y2|), which has the same property. It's neighborhoods are square with edges parallel to the x and y axes.