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Math Help - Open, closed or neither?

  1. #1
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    Open, closed or neither?

    How do I prove that the set


    A=\left\{ \left(a,b\right)=\in\mathbb{R}^{2}:\:0\leq x<3,\:0\leq y\leq5\right\}

    is open, closed or neither? I believe it to be neither, because of the x<3 condition.

    What would the closure, interior and boundary of the set A be?

    Thanks in advance.
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  2. #2
    Super Member TheChaz's Avatar
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    Re: Open, closed or neither?

    How do you prove it? Using the definitions!

    It's not open, since - for example - the point (0, 0) doesn't have any open balls/neighborhoods contained in A.

    What about it being closed? It might help to find the closure first...
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  3. #3
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    Re: Open, closed or neither?

    Thanks for your post TheChaz. However, this doesn't really help me.

    I can see "intuitively" why it is not open or closed. But it's the proof that has got me.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Re: Open, closed or neither?

    Quote Originally Posted by Cairo View Post
    How do I prove that the set


    A=\left\{ \left(a,b\right)=\in\mathbb{R}^{2}:\:0\leq x<3,\:0\leq y\leq5\right\}

    is open, closed or neither? I believe it to be neither, because of the x<3 condition.

    What would the closure, interior and boundary of the set A be?

    Thanks in advance.
    I will answer your second question, maybe that will help you find the answer:

    The closure is of course, the closed rectangle: \left \{ (x,y) \in \mathbb R ^2 ~:~0 \le x \le 3,~0\le y \le 5 \right \}

    The interior is: \left \{ (x,y) \in \mathbb R ^2 ~:~0 < x < 3,~0 < y < 5 \right \}

    The boundary is: \left \{ (x,y) \in \mathbb R ^2 ~:~x = 0 \text{ or }x = 3,~\text{with } 0 \le y \le 5 \right \} \cup \left \{ (x,y) \in \mathbb R ^2 ~:~y = 0 \text{ or } y = 5, \text{ with } 0 \le x \le 3 \}

    Does that help you "solve" the problem? What are the definitions of "open", "closed" you are using?
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  5. #5
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    Re: Open, closed or neither?

    Thanks for this Jhevon.

    So the set A is not open since (0,0) in A, but there is no closed ball centred at (0,0) with radius greater than zero that is contained in A. Also, A is not closed since {a_k = (3-1/k , 5)} is a sequence of points in A, but x_k tends to 3 (which is not a member of A) as k tends to infinity.

    Having used the definitions of open and closed sets, would this be classed as a proof, or does it need to be more rigorous?
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  6. #6
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    Re: Open, closed or neither?

    it would be better to say that {a_k} tends to (3,5), which is not a member of A. (3,5) is a "near point" of A, but it is not in A. a closed set contains all of its "near points".

    (note: the term "near point" may appear to be vague. it's actually possible to define a "near point" rigorously. a point a is near A, written a←A, if every neighborhood of a has a non-empty intersection with A. note the similarity with "limit point", which is the same except for using deleted neighborhoods of A

    the concept of "near point" is what many people mistakenly believe "limit points" to be, and in my opinion, makes for better visualization.

    since isolated points of A (points in A that aren't limit points of A) are in A, A U {limits points of A} = A U {near points of A}.).
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  7. #7
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    Re: Open, closed or neither?

    Thanks, Deveno.

    So are you suggesting that the set A is in fact closed? The way I read your post would suggest that the set does contain all of its near points, since the sequence I chose was arbitrary.

    The sequence {a_k = (3-1/k , n)} where n runs through 0 to 5 would show that every point in this sequence is a near point.

    Or have I misinterpreted what you have said?
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  8. #8
    Super Member TheChaz's Avatar
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    Re: Open, closed or neither?

    Quote Originally Posted by Cairo View Post
    Thanks for your post TheChaz. However, this doesn't really help me.

    I can see "intuitively" why it is not open or closed. But it's the proof that has got me.
    That is the proof!
    A set is said to be open if, "for all..."
    Therefore, a set is *not* open if, "there exists..."

    So it is sufficient to find one counterexample of the claim that A is open, to prove that A is not open.
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