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:
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:
From (a) and (b) it follows :
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.
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.
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. 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:
Opened unit ball in discrete metric:
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.