Hello,
Could anybody please help me to solve some problems:
1. Prove that closure of opened ball with radius r in normed linear space is equal to closed ball with radius r, r>0.
2. Show that interior of the closed ball with radius r in normed linear space is equal to opened ball with radius r, r>0.
3. Show that 1., 2. doen't apply to metric space.
4. Prove that closure of space Y is closed vector subspace of X, if Y is vector subspace of normed linear space X.
5. Show that closure of K is closed convex subspace of X, if K is convex subspace of normed linear space X.
Thank you in advance.
The problem is that I don't have any experience with such problems, and I am more than sure, that my solutions aren't correct or the formal description of the solutions isn't correct.
Here is my attempt to solve the 1. problem:
Definition of the closure:
.
For
. (a)
It remains to check whether the points on the boundary of the U(0,r) satisfy the condition: . As for every we can find such s that:
, then
. (b)
From (a) and (b) it follows :
B(0,r)-closed ball
Please take a look what is wrong, or suggest any other way of solution. Any comments will be useful for me.
I will post the rest of my solutions later.
I think you are confusing the field with the elements of the space. r is a number, and so it makes no sense to say s-r. Here is a proof of the statement.
The by definition we have (*) and where B(r,x) is the closed ball around x of radius r. As the closure is defined to be the smallest closed superset, we have . To prove the other inclusion take . If , then by (*) we have that . Otherwise . Now consider the sequence . This sequence converges to y, and so using this fact (observe that the sequence is in , if you don't believe me take the norm of y_n-x), we can say that .
If you have any more problems (with this or other questions), post here.
Can we use a similar approach for the 2.problem in the following way?
As the int B(r,x) consists of points with neighbourhood in B(r,x), it follows that
.
If , then we can observe sequence
which converges to , but every element of the sequence isn't in the B(r,x). It means that there is no such neighbourhood of , which is in B and such points don't belong to int B(r,x). It means that (opened ball).
With the discrete metric I can prove only that 1. statement doesn't apply to the metric space. From my point of view, 2. statement applies to the metric space.
My solutions:
1. statement (1.problem).
For the discrete metric in space X we have: d (x,x)=0, d(x,y)=1, . That's why we can observe only unit balls.
(1)
.
d(x,y)=0 only when x=y=x0 only for the x_0, because there is no such sequence in U(1,x0) with limit x if . It follows:
(2)
2. statement (2.problem).
. For the discrete metric we have only one possibility for r>0, r=1. And only x0 satisfies the condition d(x,y)<1, x=y=x0. It follows:
(3)
Opened unit ball in discrete metric:
(4)
(3)=(4)
Could you please tell me, what I'm doing wrong? Thank you in advance.
I have a solution for the 4.problem. But I don't have any idea about how to solve 5. problem.
Solution for 4.pr.:
. If . As intCY is the biggest opened subspace of CT (by definition of the interior), then is smallest closed subset which includes Y, because complement of the open space is closed space.