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Thread: closed open perfect bounded

  1. #1
    No one in Particular VonNemo19's Avatar
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    closed open perfect bounded

    I'm supposed to determine whether the following sets (all subsets of $\mathbb{R}^2$) are closed, open, perfect, and /or bounded.

    A finite set.
    This is strange. Isn't any neighborhood finite? Those are open. Is the book saying any finite subset of $\mathbb{R}^2$ is closed.

    The set of all integers.
    Am I wrong to say that this set has no limit points? What does it look like in $\mathbb{R}^2$? All the points $(x,y)$ in the plane where $x,y\in\mathbb{Z}$ ?

    The set consisting of the numbers $\frac{1}{n},~~(n=1,2,3,...)$.

    The set of complex numbers.

    The segment $(a,b)$.

    I thought I had this before these examples. If maybe you could justify one of these for me it would help.
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    Re: closed open perfect bounded

    Quote Originally Posted by VonNemo19 View Post
    I'm supposed to determine whether the following sets (all subsets of $\mathbb{R}^2$) are closed, open, perfect, and /or bounded.
    A finite set.
    This is strange. Isn't any neighborhood finite? Those are open. Is the book saying any finite subset of $\mathbb{R}^2$ is closed. The set of all integers.
    Am I wrong to say that this set has no limit points? What does it look like in $\mathbb{R}^2$? All the points $(x,y)$ in the plane where $x,y\in\mathbb{Z}$ ?
    The set consisting of the numbers $\frac{1}{n},~~(n=1,2,3,...)$.
    The set of complex numbers.
    The segment $(a,b)$.
    I thought I had this before these examples. If maybe you could justify one of these for me it would help.
    A lot of this is simply a matter of being able to read plain language.
    Think about this TRUE statement: If a set $\mathscr{A}$ has no limit points then it contains all of its limit points. At first, that may seem odd but nonetheless it is true. Therefore a set with no limits points is closed.

    You ask about neighborhoods is $\mathbb{R}^2$. They are simply interiors of circles. That makes the plane a Hausdorff Space: each two points are separated. Suppose that $p~\&~q$ are points in the plane. Them the distance $\varepsilon=|p-q|>0$, then if $\delta=\frac{\varepsilon}{2}$, the circular neighborhoods $\mathscr{N}_{\delta}(p)\cap\mathscr{N}_{\delta}(q )=\emptyset$ Be sure that you understand that.

    Again that may seem counter-intuitive but important. So any neighborhood of a limit contains infinitely many points of the set. So can a finite set have a limit point?
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    Re: closed open perfect bounded

    Quote Originally Posted by VonNemo19 View Post
    I'm supposed to determine whether the following sets (all subsets of $\mathbb{R}^2$) are closed, open, perfect, and /or bounded.

    A finite set.
    This is strange. Isn't any neighborhood finite? Those are open. Is the book saying any finite subset of $\mathbb{R}^2$ is closed.
    No, neighborhoods are not, in general, finite. An open interval, and neighborhood, (a, b) has finite length (b- a) but contains and infinite number of points. More generally, in any metric space, N_p(r), the set of all points with distance from point p less than positive real number, r, has finite radius but contains an infinite number of points. When the problem says "a finite set", it is referring to a set that has a finite number of points. It is fairly easy to prove that any "singleton set", a set containing a single point, is closed. There is also a theorem that says that the union of any finite set of closed sets is closed. From those two it follows that any finite set it closed.

    The set of all integers.
    Am I wrong to say that this set has no limit points? What does it look like in $\mathbb{R}^2$? All the points $(x,y)$ in the plane where $x,y\in\mathbb{Z}$ ?
    You are correct that the set of all integers has no limit points. No, the set of all integers is not, strictly speaking, a subset of $R^2$ at all. It is the set {..., -3, -2, -1, 0, 1, 2, 3, ...}. We can "associate" the set of all integers with the set {(n, 0)} with n any integer or, for that matter, the set {(n, a)} where n is any integer and a is any single real number. (By "associate" I mean that the mapping n->(n, a) is a "homeomorphism".) Since this set "has no limit points", the statement "it contains all of its limit points", which is equivalent to the statement "if p is a limit point for the set, it is in the set, is trivially true so this is a closed set.

    The set consisting of the numbers $\frac{1}{n},~~(n=1,2,3,...)$.
    This is the set {1, 1/2, 1/3, 1/4, ...}. Do you see that it has a single limit point? Is that limit point in the set?

    The set of complex numbers.
    I take it you mean the complex numbers as a subset of them selves, not as a proper subset of any other set. A basic property of any topological space is that the entire space is both open and closed.

    [quote]The segment $(a,b)$.[quote]
    That is referred to as the "open interval" which should you give you a hint! It is the set of all real numbers strictly between a and b. Actually, whether a set is open or closed depends on what topology you are using. I presume that you mean the "usual" topology on the real numbers, the metric topology defined by d(x, y)= |x- y|.

    Let p be any point in the set (a, b). Then a< p< b. That, in turn, means that p- a> 0 and b- p> 0. How we continue depends upon which of those numbers is smaller- whether p is closer to a or b. If p- a is smaller than b- p, so p is closet to a than to b, let \delta= (p- a)/2. The neighborhood of p, N_p(\delta)= {p- \delta, p+ \delta) is completely inside (a, b) so p is an interior point of (a, b). If b- p is smaller than p- a, so p is closer to b than to a, let \delta= (b- p)/2. The neighborhood of p, N_p(\delta)= {p- \delta, p+ \delta) is completely inside (a, b) so p is an interior point of (a, b). In either case, p is an interior point of (a, b). Every point of (a, b) is an interior point so (a, b) is an open set.

    I thought I had this before these examples. If maybe you could justify one of these for me it would help.
    One difficulty here is that different texts, and teachers, can use different definitions (but equivalent) of "open" or "closed", etc. Since you showed no attempt yourself to do these we do not know which definitions you are expected to use.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Re: closed open perfect bounded

    This is the set {1, 1/2, 1/3, 1/4, ...}. Do you see that it has a single limit point? Is that limit point in the set?
    This opened my eyes to what limit points can be. I was thinking that continuity within the set was necessary to even begin to think about limits. The elements of this set are spaced apart, but any neighborhood about 0 contains points neq 0 in the set.


    that is referred to as the "open interval" which should you give you a hint...
    The way you worked this was clear and concise.

    No, the set of all integers is not, strictly speaking, a subset of $R^2$ at all
    This is what had me confused. The book says:
    "Let us consider the following subsets of $R^2$:
    (a) the set...
    (b) a finite set..
    (c)...
    ...
    (e) The set of all integers.

    Hence my confusion.

    @plato
    A finite set of points has no limit points, right?
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    Re: closed open perfect bounded

    Quote Originally Posted by VonNemo19 View Post
    This is what had me confused. The book says:
    "Let us consider the following subsets of $R^2$:
    (a) the set...
    (b) a finite set..
    (c)...
    ...
    (e) The set of all integers.
    @plato, A finite set of points has no limit points, right?
    We know that you are self-studying. That can be very good or very bad, there is no middle ground.
    I have found that topology(metric spaces is a subtopic) is the most confusing study one can attempt.

    Thus please tell us what textbook you are using.
    The reason I ask is simple: Any textbook that does not make it perfectly clear to you that the set of integers is NOT a subset of $\mathbb{R}^2$ is not worth its salt or you do not have the background to read it. Which is it?

    Now to answer the question you directed to me.
    THEOREM: In a metric space if $p$ is a limit point of a set $\mathcal{A}$ then any neighborhood $\mathscr{N}_d(p)$ must contain infinitely many points of $\mathcal{A}$.
    Now, it is not an overstatement to say: if you cannot prove that even in your sleep, if you do not understand its implications then your further study is pointless.
    But there is hope. If you work through the small textbook Elementary Theory of Metric Spaces by Robert Reisel you may be able to go on.
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Re: closed open perfect bounded

    I am studying rudin's principles of mathematical analysis, and I must say...it's not easy going. I've come to a certain level of maturity in math, and I'm ready to tackle these subjects. It's just hard. I only have myself, the internet, and books as resources. I'm not trying to go all topological right now. Its just that the chapter I'm studying is called basic topology. I think I need a different book.

    which is it?
    In this case, I think a little of both.
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    Re: closed open perfect bounded

    Quote Originally Posted by VonNemo19 View Post
    I am studying rudin's principles of mathematical analysis, the internet, and books as resources. I'm not trying to go all topological right now. Its just that the chapter I'm studying is called basic topology. I think I need a different book.
    HERE is a web-page that is helpful with Rudin.
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