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Math Help - I don't even know where to begin with this stuff...

  1. #1
    Member billym's Avatar
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    I don't even know where to begin with this stuff...

    Can somebody explain to me (in fool's english) why or why not the union of an open set and a closed set is open, closed, neither or both?

    What about Z? What about Q?

    ...

    [0,1] U [2,3] U [4,5] U ... U [2n, 2n + 1] U ...

    I assume this is an infinite collection of closed sets. Great.
    If the union of a finite collection of closed sets in R is closed... does that mean this is open?
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  2. #2
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    Quote Originally Posted by billym View Post
    Can somebody explain to me (in fool's english) why or why not the union of an open set and a closed set is open, closed, neither or both?
    Depending on the sets, the union can be any of those.

    Suppose A= (0, 1), an open set, and B= [-1,2], a closed set. Then their union is [-1, 2] which is closed.

    Suppose A= [0, 1], a closed set, and B= (-1,2), an open set. Then their union is (-1, 2) which is open.

    Suppose A= (0, 1), an open set and B= [1, 2], a closed set. Then their union is (0, 2] which is neither open nor closed.

    Suppose A= (-\infty, 1), an open set, and B= [0, \infty), a closed set. Their union is all of R which is both open and closed.

    What about Z? What about Q?
    What about them? Are you asking whether they are open or closed? In what topology? As subsets of the real numbers with the "usual" topology"?

    In that case, Z is closed and Q is neither open nor closed. Z is closed because it has NO limit points (no sequence of integers converges) and so it is correct to say that it contains all of its limit points (the set of all of its limit points is empty and the empty set is a subset of all sets). Q is not closed because it has limit points that are not rational numbers. For example, the sequence 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, 3.141592, ..., where each number is \pi to one more decimal place, converges to \pi so \pi is a limit point but is not a rational number. Q is not open because every interval contains some irrational numbers. In particular, an interval centered on any rational number, no matter how small, contains some irrational numbers so that rational number is not an "interior point".
    ...

    [0,1] U [2,3] U [4,5] U ... U [2n, 2n + 1] U ...

    I assume this is an infinite collection of closed sets. Great.
    If the union of a finite collection of closed sets in R is closed... does that mean this is open?
    No. While there is a theorem that says the union of a finite number of closed sets is closed, there is no theorem that says one way or the other about the union of an infinite number of closed sets. That particular example happens to be closed because its complement, (-\infty, 0)U(1, 2)U(3,4)U...U(2n-1,2n)U... is a union of open sets and so open. But \cup_{n=1}^\infty [1/n, n] is a union of an infinite number of closed sets that is open- it is (0, \infty)- while \cup_{n=1}^\infty[1/n, 2] is a union of an infinite number of closed sets that is neither open nor closed- it is (0, 2]. And, of course \cup_{n=1}^\infty [-n, n] is a union of an infinite number of closed sets which is both open and closed- it is all of R.

    Also remember that saying a set is "not closed" does not necessarily mean it is open. It might be neither open nor closed.
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  3. #3
    Member billym's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Suppose A= (0, 1), an open set, and B= [-1,2], a closed set. Then their union is [-1, 2] which is closed.

    Suppose A= [0, 1], a closed set, and B= (-1,2), an open set. Then their union is (-1, 2) which is open.

    Suppose A= (0, 1), an open set and B= [1, 2], a closed set. Then their union is (0, 2] which is neither open nor closed.

    Suppose A= (-\infty, 1), an open set, and B= [0, \infty), a closed set. Their union is all of R which is both open and closed.
    All of this makes sense because the sets intersect (except for the third one, but there aren't any numbers between 1 and 1 so it doesn't cause me any problems).

    But what about the union of, say, (1,2) and [3,4]? The limit points 3 and 4 are contained but 1 and 2 are not. Does this mean the set is neither open nor closed?
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    Quote Originally Posted by billym View Post
    But what about the union of, say, (1,2) and [3,4]? The limit points 3 and 4 are contained but 1 and 2 are not. Does this mean the set is neither open nor closed?
    I have a suggestion for you. Why not bite the bullet so to speak?
    Just learn the basic definitions.
    What does it really mean for a set to be open? Learn examples.
    What does it really mean for a set to be closed? Learn examples.
    Also realize that a door is either open or closed; that is not true of sets.
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  5. #5
    Member billym's Avatar
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    Well I try to, but they always seem to be buried in 38 non-basic definitions. Things seem to make the most sense when I can see worked examples. I've read through a few intro-analysis 'this and thats', and obviously I know what a set is. But I can't find one example of a union of disjoint open and closed sets.

    To be fair, isn't analysis all about delighting in counter-intuitive surprises anyway?

    (Time is also a factor)
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  6. #6
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    Quote Originally Posted by billym View Post
    Well I try to, but they always seem to be buried in 38 non-basic definitions. Things seem to make the most sense when I can see worked examples. I've read through a few intro-analysis 'this and thats', and obviously I know what a set is. But I can't find one example of a union of disjoint open and closed sets.

    To be fair, isn't analysis all about delighting in counter-intuitive surprises anyway?
    What ever all that means, that has nothing to do with the questions I put to you.
    You have missed the point. Do you understand the definitions?
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  7. #7
    Member billym's Avatar
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    Like, that a closed set contains its limit points and an open set doesn't?

    I'm just wondering if (2,8] and (2,4)U[5,8] are the same in terms of being open or closed.
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  8. #8
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    Quote Originally Posted by billym View Post
    Like, that a closed set contains its limit points and an open set doesn't?
    I'm just wondering if (2,8] and (2,4)U[5,8] are the same in terms of being open or closed.
    You just have real problems with definitions.
    You need to learn how to read mathematics.
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    Quote Originally Posted by billym View Post
    Like, that a closed set contains its limit points and an open set doesn't?

    I'm just wondering if (2,8] and (2,4)U[5,8] are the same in terms of being open or closed.
    The set of limit points of (2,8] is [2,8].

    The set of limit points of (2,4)U[5,8] is [2,4]U[5,8].

    Then again, your definition of an open set is wrong.

    On the real line, a set is open iff for any x in that set, there exists some epsilon > 0 such that (x - epsilon, x + epsilon) is a subset of that set.

    A set is closed if its complement is open.

    Hope that helps.
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  10. #10
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    Quote Originally Posted by billym View Post
    Like, that a closed set contains its limit points and an open set doesn't?

    I'm just wondering if (2,8] and (2,4)U[5,8] are the same in terms of being open or closed.
    No.

    Plato has said this already, but you seem to have missed it: Some sets are neither open nor closed. In fact, some sets (e.g. R) are both!

    Here are two very straightforward definitions:

    A set S\subseteq\mathbb{R} is closed if and only if for each accumulation point (or limit point) x we have x\in S.

    Also:

    A set T\subseteq\mathbb{R} is open if and only if for each x\in T there is a neighborhood Q of x with Q\subseteq T.

    You may also benefit from the following reminders:

    A set Q\subseteq\mathbb{R} is a neighborhood of x\in\mathbb{R} if and only if there is \epsilon>0 with (x-\epsilon,x+\epsilon)\subseteq Q.

    Also:

    If S\subseteq\mathbb{R}, then x is an accumulation point (or limit point) of S if and only if x\in\mathbb{R} with every neighborhood of x containing infinitely many elements of S.

    Please note that this last definition is not necessarily correct outside the context of real analysis. In other branches you must use a more rigorous definition, but for our purposes the above will suffice.

    Okay, so now you are armed with these four definitions. Use them carefully.

    Consider for example \mathbb{R}. It is trivial that \mathbb{R}\subseteq\mathbb{R}. Furthermore, since accumulation points of \mathbb{R} are in \mathbb{R} by definition, we know that \mathbb{R} is a closed set.

    But wait!

    Notice also that for any accumulation point x\in\mathbb{R} and \epsilon>0, we have (x-\epsilon,x+\epsilon)\subset\mathbb{R}. So \mathbb{R} is an open set, too!

    Thus we have an example of a set which is both open and closed. You can also find examples of a set which is neither open nor closed (e.g. (0,1]).

    Hopefully that helps.
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  11. #11
    Member billym's Avatar
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    Right, so (2,4)\cup[5,8] is not open because the points x = 5 and x = 8 are contained in the union, but every neighborhood of these points is not contained in the union.

    It is also not closed because its complement (-\infty,2]\cup[4,5)\cup(8,\infty) is not open at 2 and 4.

    ?
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    Quote Originally Posted by billym View Post
    Right, so (2,4)\cup[5,8] is not open because the points x = 5 and x = 8 are contained in the union, but every neighborhood of these points is not contained in the union.

    It is also not closed because its complement (-\infty,2]\cup[4,5)\cup(8,\infty) is not open at 2 and 4.

    ?
    Well, I would not say that "not open at 2 and 4". It is not open because the points 2 and 4, which are in the set, are not interior points.

    In a general "metric" space, we define an " \epsilon neighborhood of point x" (for \epsilon> 0) to be the set of all points in the space whose distance from x is less than \epsilon- that is, \{ y | d(x, y)< \epsilon\}.

    For the real numbers, the "usual metric" is d(x,y)= |x- y|. For the plane, R^2 it is d((x_1,y_1), (x_2, y_2))= \sqrt{(x_2-x_1)^2+ (y_2- y_1)^2}, the usual distance between two points calculation.

    A point x is said to be an "interior point" of set A if and only if there is some \epsilon neighborhood of x that is a subset of A.

    A set is "open" if all of its points are interior points.

    A set is "closed" if its complement is open.


    Here is a slight variation on those definitions that I like.

    "Interior point" is defined as above.

    A point, x, is called an "exterior point" of set A if it is an interior point of the complement of A.

    A point, x, is called a "boundary point" of set A if it is neither an interior point nor an exterior point of A.

    Notice that all points in the space can be classified as exactly one of those three with respect to set A. An "interior point" of A must be in A so it cannot be in the complement of A- no interior point of A is also am exterior point of A and vice-versa. And, of course, the definition of "boundary point" as "neither an interior point nor an exterior point" means a boundary point cannot be either an interior point nor an exterior point.

    Further, the exterior points of A are, by definition, the interior points of the complement of A and, since complement is "duel" (the complement of the complement of A is again A) the interior points of A are the exterior points of the complement of A. And any set A and its complement have exactly the same boundary points.

    Now, we can define a set to be open if and only if it contains NONE of its boundary points and define a set to be closed if and only if it contains all of its boundary points. Because a set and its complement have the same boundary points, it follows that the complement of an open set is closed and vice-versa.

    Of course, a set may be neither open nor closed. If contains some but not \all of its boundary points, it is neither open nor closed.

    It might sound like a set could not be both open and closed. That would require that it contain all of its boundary points and contain none of its boundary points! How is that possible?

    Very easily- as set that has no boundary points contains all of its boundary points (trivially) and, at the same time, contains no boundary points. For the real numbers the only sets that have no boundary points are the empty set (All numbers are exterior points for the empty set. It has no interior points and no boundary points.) and the set of all real numbers (All numbers are interior points for the set of all real numbers. It has no exterior points and no boundary points.).

    It can be shown that if a space is connected the empty set and the entire set are the only subsets that are both closed and open. But if a set is not connected, all of its "components" (a connected subset that is not properly contained in any larger connected set) are both open and closed.
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    Quote Originally Posted by billym View Post
    Right, so (2,4)\cup[5,8] is not open because the points x = 5 and x = 8 are contained in the union, but every neighborhood of these points is not contained in the union.

    It is also not closed because its complement (-\infty,2]\cup[4,5)\cup(8,\infty) is not open at 2 and 4.

    ?
    For the second one, recall that
    A set S is not closed \iff its complement is not open \iff \exists x \in  ^cS s.t.  \exists some  \delta > 0 s.t.  (x-\delta, x+\delta) is not a subset of ^cS.

    So the correct way to put it would be that 2, 4 are two such points. Of course this metric applies only to \Re. For \Re^2 HallsofIvy has explained it succinctly, and the metric would differ for other spaces.
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  14. #14
    Member billym's Avatar
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    Boundary points:

    If I have a subset: {1/2, 1/3, 1/4, ... 1/n}, then all of its points are boundary points because none of its elements are interior points of the subset(because they all have nbhd's that are not in the subset), nor are they interior points of the subset's complement (obviously).

    Correct?
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    Quote Originally Posted by billym View Post
    Boundary points:

    If I have a subset: {1/2, 1/3, 1/4, ... 1/n}, then all of its points are boundary points because none of its elements are interior points of the subset(because they all have nbhd's that are not in the subset), nor are they interior points of the subset's complement (obviously).

    Correct?
    Clearly, as you noted each element \frac{1}{k} of S=\left\{\frac{1}{n}:n\in\mathbb{N}\right\} has no other elements of S in the neighborhood of radius \frac{1}{2(k+1)}. Every point of that neighborhood is in \mathbb{R}-S except \frac{1}{k} itself. Clearly then \frac{1}{k}\in\partial S and since \frac{1}{k} was arbitrary every point of S is in \partial S
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