# Problem with set of inside proofs

1. Mar 27, 2013

### trenekas

I have a problem. Here is the task. Need to prove all these claims:

I was able to prove only one and half of them.

b) Let A$\subset$B and x $\in$ A°. It mean that there is ε>0 that Oε(x)$\subset$A. And it mean that Oε(x)$\subset$A$\subset$B. And that mean that x$\in$B, what I need to prove.

c) (A°)°$\subset$A° it follows from a. if i could prove that :D
How to prove A°$\subset$ (A°)° i dont know. tried to apply definition but nothing goes on.

Can you give me some advise which way i need to go?

Last edited: Mar 27, 2013
2. Mar 27, 2013

### Zondrina

Part (a) says that the set of interior points of a set is a subset of the original set. This one really shouldn't be too difficult.

Suppose $x \in A°$. You want to somehow show that $x \in A$. Think about delta neighborhoods around x.

3. Mar 27, 2013

### trenekas

Yeah. a) i proved but wasnt sure if it correct. but second part of c, d and e are to difficult. I have no idea from what need to start

4. Mar 27, 2013

### trenekas

And also i want you ask very similar question. I think i proved this, but want to know if it correct or not.

The task: Need to prove that open rectangle ([a,b]):={x$\in$R^d: a<x<b}=(a1,b1)x...x(ad,bd) is open set.
I think i can prove that every interval of this set is open and that means that all set is open? Correct? :) if that is true so all that intervals R\(a1,b1) ... R\(ad,bd) is closed. And according to one theorem which i found in my teacher book it mean that intervals (a1,b1), (a2,b2)... (ad,bd) are open.

Do you think this way of thinking is correct or not? :) Thanks!

5. Mar 27, 2013

### jbunniii

Do you know the characterization that $A^\circ$ is the union of all open sets contained in $A$? Equivalently, it is the largest open subset of $A$ (in the sense of containment). This will easily allow you to conclude (a) and (b).

Another useful fact which follows from this is that $A = A^\circ$ if and only if $A$ is open. This immediately gives you (c).

6. Mar 27, 2013

### Zondrina

I believe it would be informative if we knew what class this was for; so we know what information you may have access to.

7. Mar 27, 2013

### jbunniii

It is certainly true that each $(a_k, b_k)$ is an open subset of $\mathbb{R}^1$, but you will have to do more work if you want to show that $(a_1, b_1) \times \ldots \times (a_d, b_d)$ is an open subset of $\mathbb{R}^d$.

Here is a hint to get you started. Choose an arbitrary element $x = (x_1, \ldots, x_d) \in (a_1, b_1) \times \ldots \times (a_d, b_d)$. You need to show that $x$ is an interior point. That means you have to find a $\delta$ such that the open ball centered at $x$ with radius $\delta$ is contained in $(a_1, b_1) \times \ldots \times (a_d, b_d)$. To find such a $\delta$, use the fact that $x_k$ is an interior point of $(a_k, b_k)$ for each $k$.

8. Mar 27, 2013

### trenekas

Thanks once more:)

9. Mar 27, 2013

### trenekas

Dont know how well I understand your hint. so now ill write the ending of proof :) if something is wrong please tell me.
We know that x1 is (a1,b1) interior point. so there is ε1>0 that ||a1-b1||<ε1
...
...

let $\delta$= min{$\epsilon1$, $\epsilon2$,...$\epsilon d$}
And that mean that open ball with radius $\delta$ contained in (a1,b1)x...x(ad,bd)

correct? :)

Last edited: Mar 27, 2013
10. Mar 27, 2013

### jbunniii

It is true, but why? If you take an arbitrary point $y$ in the same open ball (i.e. such that $\|x - y\| < \delta$), can you show that $y \in (a_1,b_1)\times \ldots \times (a_d,b_d)$?

11. Mar 27, 2013

### trenekas

all the same but ε1>0 and |(|a1-b1|)-x1|<ε1
...
...

let $\delta$= min{$\epsilon1$, $\epsilon2$,...$\epsilon d$}

And that mean that open ball with radius $\delta$ contained in (a1,b1)x...x(ad,bd)

now is good?

Last edited: Mar 27, 2013
12. Mar 27, 2013

### jbunniii

I think these inequalities are still wrong. For each $k$, you want both of the following to be true:
$$a_k < x_k - \epsilon_k$$
$$x_k + \epsilon_k < b_k$$
You can achieve this by choosing $0 < \epsilon_k < \min(x_k - a_k, b_k - x_k)$.

13. Mar 27, 2013

### trenekas

yes. my bad once more. If $\epsilon$ will be a little less than $\min(| x_k - a_k |,| b_k - x_k |)$ so all that would be correct when $\delta$ equal to $\epsilon$

Last edited: Mar 27, 2013
14. Mar 27, 2013

### jbunniii

Assuming you take the minimum over all $k$ (i.e., $\delta_k = \min(|x_k - a_k|, |b_k - x_k|)$, and $\delta = \min_k \delta_k$), it will work, but you still have to prove that.

15. Mar 27, 2013

### trenekas

how to do that? i think if we found $\delta$ when all is good.

I dont know or my reasoning is good but i think that, when we found a $\delta$ as you described when for every x will be neighbourhood which will be subset of our first set. And that mean that every x is interior point of first set.

if that not enough i give up. Dont know how this prove...

Last edited: Mar 27, 2013
16. Mar 27, 2013

### trenekas

i want to add that $||x-a||$>$\delta$ and also $||x-b||$>$\delta$ so from here i draw conclusion in my previous post.Its good? Please someone help me!!! ))

17. Mar 27, 2013

### jbunniii

If $B_\delta(x)$ is the open ball of radius $\delta$ centered at $x$, and $y = (y_1,\ldots,y_d)$ is any point in $B_\delta(x)$, then you need to show that $y \in (a_1, b_1) \times \ldots \times (a_d, b_d)$. In order to do this, you must show that $a_k < y_k < b_k$ for all $k$.

So suppose $y \in B_\delta(x)$. By definition, that means that the distance between $x$ and $y$ is less than $\delta$, so
$$\sqrt{(y_1 - x_1)^2 + \ldots + (y_d - x_d)^2} < \delta$$
You have to prove that this inequality implies that $a_k < y_k < b_k$ for all $k$. You will of course have to make use of the way we defined $\delta$, namely $\delta_k = \min(|x_k - a_k|, |b_k - x_k|)$, and $\delta = \min_k \delta_k$.

18. Mar 27, 2013

### trenekas

ok. thank you jbunnii!!! tomorow i rethinking all what you say because my brain now is not capable to think :) need some rest.