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Math Help - regular open/closed sets

  1. #1
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    regular open/closed sets

    I need to show that a set B is regular open if and only if its complement is regular closed.

    => Assume B is regular open => B = Int (cl B)

    <= Assume C(B) is regular closed => B = cl (Int (B))

    I know that C(Int B)= cl (C(B)) and Int (B)= C(cl C(B)) [The int B = the complement of the closure of the complement of B]

    I'm just not sure how to relate those to what I have above.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by EricaMae View Post
    I need to show that a set B is regular open if and only if its complement is regular closed.

    => Assume B is regular open => B = Int (cl B)

    <= Assume C(B) is regular closed => B = cl (Int (B))

    I know that C(Int B)= cl (C(B)) and Int (B)= C(cl C(B)) [The int B = the complement of the closure of the complement of B]

    I'm just not sure how to relate those to what I have above.
    Ok So let X be a metric space, and let E\subset{X}, and let E^c represent the complement of E. Now assume that x is a limit point of E^c but x\notin{E^c}\implies{x\in{E}}. Since by definition of a limit point \exists{p}\in{E^c}\in{N_r{(x)}}, but this leads to a contradiction because x\in{E} and E is open so that means that \exists{r>0}\backepsilon{N_r(x)\subset{E}}. So in other words since any neighborhood of a limit point of E^c must contain another point of E^c, it cannot be an interior point of E thus all limit points of E^c are in E^c, thus it is closed .

    I hope that made sense
    Last edited by Mathstud28; November 18th 2008 at 03:09 AM.
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  3. #3
    Moo
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    Quote Originally Posted by Mathstud28 View Post
    Ok So let X be a metric space, and let E\subset{X}, and let E^c represent the complement of E. Now assume that x is a limit point of E^c but x\notin{E^c}\implies{x\in{E}}. Since by definition of a limit point \exists{p}{\color{red}\in{N_r{(x)}} \subset{E^c}}, but this leads to a contradiction because x\in{E} and E is open so that means that \exists{r>0}\backepsilon{N_r(x)\subset{E}}. So in other words since any neighborhood of a limit point of E^c must contain another point of E^c, it cannot be an interior point of E thus all limit points of E^c are in E^c, thus it is closed .

    I hope that made sense
    Okay, I don't know what the OP understood to this, nor do I know if it's just me being picky.

    But your proof is not clear. Don't try to make it shorter. x is a limit point of E^c but x \not \in E^c ? Why ? Shouldn't you say "Let's assume x \not \in E^c" ?

    And I think at least that red part should be changed. It is no sense to say that a set "belongs" to a set (in that case).

    Also, I've never been fond of «"blabla" because there is a contradiction if». It does not follow a logical path. Make your assumptions, state clearly what you suppose. List the properties that would come from these assumptions and then conclude by saying there is a contradiction.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Okay, I don't know what the OP understood to this, nor do I know if it's just me being picky.
    But your proof is not clear. Don't try to make it shorter. x is a limit point of E^c but x \not \in E^c ? Why ? Shouldn't you say "Let's assume x \not \in E^c" ? \color{red}\text{I did, I said lets assume its a limit point of }E^c\text{ but not in }E^c

    And I think at least that red part should be changed. It is no sense to say that a set "belongs" to a set (in that case). \color{red}\text{What you have put in red I did not have in my post Moo}

    Also, I've never been fond of «"blabla" because there is a contradiction if». It does not follow a logical path. Make your assumptions, state clearly what you suppose. List the properties that would come from these assumptions and then conclude by saying there is a contradiction.
    By definition I proved that a limit point of E^c could not be an element of E because it would contradict E being open (since every point of an open set must be an interior point), So since all limit points of E^c can not be in E they must be in E^c because they are compliments, and the defintion of a closed set is a set who contains all its limit points. And I wrote this post with definitions and references specifically for the OP. I wrote it for them, not other people. So sorry if I made references to things that others are not well-versed in. I did not want to write a formal proof and do the homework for them, I wanted to give an outline. Later I will come back and write a formal proof with definitions. Thank you Moo, I appreciate when people call me out on things (even though here I think I was ok) because its forces me to get better.
    P.S. you misquoted me somehow.
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  5. #5
    Moo
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    \text{I did, I said lets assume its a limit point of }E^c\text{ but not in }E^c
    Okay puuuh that was not clear ^^
    You really shouldn't use the \implies arrows this way. Like you know "Hence x is in E"
    Sorry for that !

    \text{What you have put in red I did not have in my post Moo}
    That's normal, I changed it to the form it should be. Once again, it's a matter of simplifying with notations that is disturbing


    The other red part is not significant anymore, since the whole problem was in that first implication
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  6. #6
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    EricaMae, what do YOU mean by a "regular" open set and "regular" closed set? The previous replies talked about proving that the complement of an open set is closed but did not address the "regular" set part.
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    EricaMae, what do YOU mean by a "regular" open set and "regular" closed set? The previous replies talked about proving that the complement of an open set is closed but did not address the "regular" set part.
    My book uses regular open and open interchangably, so I assumed that he meant that. Of course he should still respond if not. But here is the proof in a more clear form that I said I would write, once again sorry for the last one not being as clear; I was in a rush.


    I will attempt to show that if E is an open set then E^c is a closed set. Before we start let us get our notation straight. In the following proof E^{c} is the "compliment" of E in other words E^c\equiv\left\{x:x\notin{E}\right\}, \chi is a metric space, and N_r(p) represents the neighborhood of p with radius r. Lastly, to let up any notation confusion let q be a non-descript point of E^c such that [tex]q\ne{p}.

    1. Ok, so let us assume that the above statement is false and if E is open that E^c is not closed. Because E^c is not closed by definition there exists a point p such that p is a limit point of E^c but p\notin{E^c}, but if p\notin{E^c}\implies{p\in{E}} by the definition of E^c.

    2. Now by definition of limit points the point p has the following charcterstic: \forall{r>0}~\exists{q}\in{N_r(p)}, in other words, for each neighborhood of p there exists in that neighborhood another element of E^c

    3. Now by definition of closed sets every point x of E must satisfy: \exists{r>0}\backepsilon{N_r(x)}\subset{E}. In other words each point in E must have a neighborhood that is a subset of E.

    4. Now here we run into the contradiction. This comes from the fact that if we attempt to apply both 3 and 4 to the point p, because EVERY neighborhood of p must contain elements of E^c( because it is a limit point of E^c) but at the same time p must have a neighborhood that is a subset of E (because it is an element of E and E is closed), but since E\cap{E^c}=\varnothing this is a contradiction.

    5. So since we have shown that any limit point of E^c cannot be in E and by definition of the compliment it must then be in E^c we have shown that every limit point of E^c is in E^c which is the definition of a closed set. \blacksquare
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  8. #8
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    The previous two parts of the problem state the following.

    Consider the real line with the standard metric topology. An open subset B of R^1 is said to be a "regular open" set if B is the interior of its closure.

    a. Show that (0, 1/2) U (1/2, 1) is not a regular closed set.

    I have the answer to this.

    b. A closed subset of R^1 is said to be regular closed set if it is the closure of its interior. A closed subset A of R^1 is nowhere dense if Int (cl A)=nulset. Show that a nowhere dense closed set is not a regular closed set.

    I"m a little confused on this part.

    c. Show that a set B is regular open if and only if C(B) is regular closed.

    SO this is why I was thinking that I should use the following laws:

    1. Int C(B)=C(cl B) [Interior of the complement=complement of the closure]
    2. cl C(B)= C(Int (B))

    because my definition as given in the problem of regular open is:
    B=Int(cl B) (the interior of the closure of B)
    and my definition as given in the problem of regular closed is :
    B= cl (Int B) [the closure of the interior of B)

    I tried to prove part 3 using the above rules.
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  9. #9
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    I think I have come to some type of proof on the last part of this problem.

    => Assume B is regular open => B= Int (cl B) => C(B)= C(Int (cl B)) = C( cl B) = cl (Int C(B)) => C(B) = cl (Int C(B)) => C(B) is regular closed.

    <= Assume C(B) is regular closed => C(B)= cl (Int C(B)) => C(C(B))= C(cl (Int C(B)) = C(cl C( cl B)) = Int (cl B) => B= Int (cl B).

    Is this right?
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