A. Let (X,\ d) be a metric space. Define a

William Curry

William Curry

Answered question

2021-12-21

A. Let (X, d) be a metric space. Define a diameter of a subset A of X and then the diameter of an open ball with center at x0 and radios r>0
B. Use (A) to show that every convergence sequence in (X, d) is bounded.

Answer & Explanation

enlacamig

enlacamig

Beginner2021-12-22Added 30 answers

Step 1
A) Let (X, d) be a metric space A be a subset of x.
We define diameter of A to be
Sup {d, (x, y)x, yA}
Let  be an open boll with center at x0 and radios r>0
Therefore, diameter of =2r
We choose x, y to be diametrically opposite two points whose midpoint is x0
Next d(x, y) at fains 2r.
Hence, for any x, y
d(x, y)d(x, x0)+d(x0, y)
r+r=2r
 diam()=2r
Step 2
B) Let (xn)nN be a sequence which is convergent in (x, d)
We assume that, it converges to xx
Choose =1
Then NN
s.t h d(xn, x)<1
Therefore, we have
N1=Sup{d(x, x1),d(x, x2),d(x, xn1),1} 
Then for any xn in the sep
d(xn, x)<1
Thus, (xn)nN is bounded

Cassandra Ramirez

Cassandra Ramirez

Beginner2021-12-23Added 30 answers

Step 1
Consider the discrete metric d on a set X:
d(x, y)={0 ifx=y1 ifxy
Consider the ball of radius r=12 centered at x
Then B(x, r)={x}
Now by definition, diam
A=sup{d(a,b):a,bA}
Applying it to our case where
A=B(x,r)
we have diameter of A equal to 0
nick1337

nick1337

Expert2021-12-28Added 777 answers

A fairly common example that isn't completely trivial is the British Rail metric on R, where 
d(x, y)=|x|+|y|
(or 0 for x=y), so B(x, r)
Has diameter 0 for x<r,
diameter r for x<r<2x and diameter
2(r-x)
for
r>2x

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