# Thread: closure, interior, convex subspace in normed linear space

1. ## closure, interior, convex subspace in normed linear space

Hello,
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.

2. Originally Posted by saxash
Hello,
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.

Help or do? This is quite long, so include attempts and point out where you are stuck.

3. 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:
$\overline{U(0,r)}=\{x:inf\{\|x-s\|\}=0,s\in U(0,r)\}$.
For $s:\|s\|
$\{x:\|x\|. (a)
It remains to check whether the points on the boundary of the U(0,r) satisfy the condition: $inf\{\|r-s\|\}=0, \|s\|. As for every $\varepsilon>0$ we can find such s that:
$0<\|r-s\|<0+\varepsilon$, then $inf\{\|r-s\|\}=0\Rightarrow$
$\{x:\|x\|=r\} \subset \overline{U(0,r)}$. (b)
From (a) and (b) it follows :
$\overline{U(0,r)}=\{x:\|x\|\leq r\}=B(0,r)= \{x:\|x\|\leq r\}$ 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.

4. 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 $U(r,x)\subset \overline{U(r,x)}$ (*) and $U(r,x)\subset B(r,x)$ 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 $\overline{U(r,x)}\subset B(r,x)$. To prove the other inclusion take $y \in B(r,x)$. If $||x-y||, then by (*) we have that $y\in \overline{U(r,x)}$. Otherwise $||y-x||=r$. Now consider the sequence $y_n=\frac{n-1}{n}y+\frac{1}{n}x$. This sequence converges to y, and so using this fact (observe that the sequence is in $U(r,x)$, if you don't believe me take the norm of y_n-x), we can say that $\inf_{s\in U(r,x)}||y-s||=0$.

If you have any more problems (with this or other questions), post here.

5. Thank you for your help. I have a question regarding the sequence you have used in the proof. Could you please explain in more detail, how it is possible to show that the sequence is in U(r,x)?

6. Originally Posted by saxash
Thank you for your help. I have a question regarding the sequence you have used in the proof. Could you please explain in more detail, how it is possible to show that the sequence is in U(r,x)?
Yes sure, it is quite simple, show that the sequence is a distance < r from x;
$
||y_n-x||=||\frac{n-1}{n}y+\frac{1}{n}x-x||=||\frac{n-1}{n}(y-x)||=\frac{n-1}{n}||y-x||=\frac{n-1}{n}r$

So by the definition of a ball, all the sequence is in the ball.
(Going backwards you can see how I came up with the sequence)

I hope this is clear.

7. 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
$\\ \{y\in B: \|y-x\|.
If $y: \|y-x\|=r$, then we can observe sequence
$y_n=\frac {n+1}{n}y-\frac{1}{n}x$ which converges to $y\in B(r,x)$, but every element of the sequence isn't in the B(r,x). It means that there is no such neighbourhood of $y:\|y-x\|=r$, which is in B and such points don't belong to int B(r,x). It means that $int B(r,x)=\{y\in B: \|y-x\|(opened ball).

8. Originally Posted by saxash
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
$\\ \{y\in B: \|y-x\|.
If $y: \|y-x\|=r$, then we can observe sequence
$y_n=\frac {n+1}{n}y-\frac{1}{n}x$ which converges to $y\in B(r,x)$, but every element of the sequence isn't in the B(r,x). It means that there is no such neighbourhood of $y:\|y-x\|=r$, which is in B and such points don't belong to int B(r,x). It means that $int B(r,x)=\{y\in B: \|y-x\|(opened ball).
Yes that argument woks. You might want to mention that the highlighted bit implies the other inclusion, viz. $\mbox{int}(B(r,x))\subset U(r,x)$.

If you want a hint on 3) consider the natural numbers with the discrete metric.

9. 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, $x\neq y$. That's why we can observe only unit balls.
$B(1,x_0)=\{x:d(x,x_0) \leq r\}=X$ (1)
$\overline{U(1,x_0)}=\{x:inf\{d(x,y)\}=0, y\in U(1,x_0)\}$.
d(x,y)=0 only when x=y=x0 $\Rightarrow inf\{d(x,y)\}=0$ only for the x_0, because there is no such sequence in U(1,x0) with limit x if $d(x,x0)=1$. It follows:
$int B(1,x_0)=x_0$ (2)
$(1) \neq (2)$

2. statement (2.problem).
$int B(1,x_0)=\{x:\exists r>0:y \in B(1,x_0) \ if \ d(x,y). 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:
$intB(1,x_0)=x_0$ (3)
Opened unit ball in discrete metric:
$U(1,x_0)=\{x:d(x_0,x)<1\}=\{x:d(x_0,x)=0\}=x_0$ (4)
(3)=(4)

Could you please tell me, what I'm doing wrong? Thank you in advance.

10. Originally Posted by saxash
Could you please tell me, what I'm doing wrong? Thank you in advance.
Classic example of over complicating a proof. You just need a counterexample. Don't assume X is a general space, take N. For both 1. and 2. take r=1

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Notice that {2} is closed, and {1,2,3} is open. So the interior of B is B, and the closure of U is U.

11. Could you please explain, why {1,2,3} is open?

12. Originally Posted by saxash
Could you please explain, why {1,2,3} is open?
Because it is the open ball of radius 1.5 around 2. Every set in the discrete topology/metric is open and closed.

13. 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.:
$\overline{Y}=\{x:inf\{\|x-y\|\}=0, \ y\in Y\}$. If $y \notin Y \Rightarrow y \in intCY \Rightarrow \overline{Y}= CintCY$. As intCY is the biggest opened subspace of CT (by definition of the interior), then $\overline{Y}$ is smallest closed subset which includes Y, because complement of the open space is closed space.

14. Pick $x,y \in \overline{K}$ and $t\in [0,1]$, then there are two sequences $\{x_n\}_{n\geq1}, \{y_n\}_{n\geq1} \subset K$ such that $x_n \rightarrow x, y_n\rightarrow y$, so we have for each n $t x_n+(1-t)y_n \in K$ by convexity of K. As the closure of K contains all the limit points of K, tending to the limit we see that $t x+(1-t) y \in \overline{K}$.