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Problem with set of inside proofs

  1. Mar 27, 2013 #1
    I have a problem. Here is the task. Need to prove all these claims:
    6gle7n.jpg

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

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

    c) (A°)°[itex]\subset[/itex]A° it follows from a. if i could prove that :D
    How to prove A°[itex]\subset[/itex] (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. jcsd
  3. Mar 27, 2013 #2

    Zondrina

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    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.
     
  4. Mar 27, 2013 #3
    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
     
  5. Mar 27, 2013 #4
    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[itex]\in[/itex]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!
     
  6. Mar 27, 2013 #5

    jbunniii

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    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).
     
  7. Mar 27, 2013 #6

    Zondrina

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    I believe it would be informative if we knew what class this was for; so we know what information you may have access to.
     
  8. Mar 27, 2013 #7

    jbunniii

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    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##.
     
  9. Mar 27, 2013 #8
    Thanks once more:)
     
  10. Mar 27, 2013 #9
    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
    ...
    ...
    there is εd>0 that ||ad-bd||<εd

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

    correct? :)
     
    Last edited: Mar 27, 2013
  11. Mar 27, 2013 #10

    jbunniii

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    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)##?
     
  12. Mar 27, 2013 #11
    Oh. I made a mistake.

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

    let [itex]\delta[/itex]= min{[itex]\epsilon1[/itex], [itex]\epsilon2[/itex],...[itex]\epsilon d[/itex]}

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

    now is good?
     
    Last edited: Mar 27, 2013
  13. Mar 27, 2013 #12

    jbunniii

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    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)##.
     
  14. Mar 27, 2013 #13
    yes. my bad once more. If [itex]\epsilon[/itex] will be a little less than ##\min(| x_k - a_k |,| b_k - x_k |)## so all that would be correct when [itex]\delta[/itex] equal to [itex]\epsilon[/itex]
     
    Last edited: Mar 27, 2013
  15. Mar 27, 2013 #14

    jbunniii

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    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.
     
  16. Mar 27, 2013 #15
    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
  17. Mar 27, 2013 #16
    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!!! :))))
     
  18. Mar 27, 2013 #17

    jbunniii

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    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##.
     
  19. Mar 27, 2013 #18
    ok. thank you jbunnii!!! tomorow i rethinking all what you say because my brain now is not capable to think :) need some rest.
     
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