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Boundness confusion

  1. Nov 24, 2011 #1
    My professor introduced to us this concept, but when she was explaining it, I don't think she even knows what she is talking about. So I've come to ask people who know their stuff to help.

    I was told that on R2, the inequalities

    (1) [tex]x > 0, y > 0[/tex]

    Is open and unbounded

    (2) [tex]x \geq 0, y \geq 0[/tex]

    Is closed and unbounded

    But this doesn't make sense to me. how could this be closed and unbounded? Clearly, x and y will never reach any finite value.

    A boundary region is which one that doesn't stretch to infinity in anyway, but a closed region is one where it contains it boundary. [tex]x \geq 0, y \geq 0[/tex] goes to infinity, so it isn't bounded, but how does this "contain" infinity?
  2. jcsd
  3. Nov 24, 2011 #2
    What definition of the closed set did you use?

    [tex]\{(x,y);\,x \geq 0, y \geq 0\} [/tex] being closed has nothing to do with it containing infinity. Simply, because the R^2 does not contain it.

    One usable definition of the closed set may be that the limit of any converging sequence of points from
    the closed set does belong to the set.
  4. Nov 24, 2011 #3


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    Given a set, S, the closure of that set, [itex]\overline{S}[/itex], is the set of limit points of S. That is, take any Cauchy sequence from S. Because the reals are a complete metric space, every Cauchy sequence has a limit point. That entire collection of limit points is the closure.

    The boundary is the set [itex]\overline{S} \setminus S[/itex].
    A set is closed if [itex]S = \overline{S}[/itex].

    Let's do this in one dimension. Let S be all of the real numbers. Is it closed? Answer is yes. Take any Cauchy sequence. That sequence must have a limit which is a real number. So the limit points of S is the reals. Hence it is closed.

    Boundedness is a separate concept from closed.
    Further a set can be open and closed at the same time. Or it can be neither.
  5. Nov 26, 2011 #4
    Big language here....

    I just started on Cauchy sequence, but I know that converging limits to real numbers on R is Cauchy. Infinity isn't a real number, so that one-dimension analogy doesn't work

    I don't know what that bar over S means
  6. Nov 26, 2011 #5


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    Then you answer the question that FroCho asked. What definition of "closed set" are you using?

    And pwsnafu told you what that "bar over S" means:
    "Given a set, S, the closure of that set, [itex]\overline{S}[/itex], is the set of limit points of S".

    Although that definition is not quite correct. The closure of S is the set of limit points of S union S itself. The two are different if S has "isolated" points. If [itex]S= (0, 1)\cup\{2\}[/itex], the "limit points" are in [0, 1] but the closure includes 2 also.

    Another definition of "closure of S" is "the smallest closed set" that contains S. Yet another is "the intersection of all closed sets that contain S".

    So, once again, what definition of "closed set" are you using?
  7. Nov 26, 2011 #6

    I am not sure if this answers the definition, I've always thought being closed means it contains it's "closes" it's boundary. Infinity can never be closed.

    For [tex]\geq[/tex], it's like having a door open in a room and stretches infinity far.

    For >, it's like having a line drawn around the room and the door open stretching infinity far.

    You see where I am getting at? The door is open at all times

    I also still dont' know what closure sets are because I don't know what metric spaces are and wikiapedia opened up more questions...
    Last edited: Nov 26, 2011
  8. Nov 27, 2011 #7


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    Nice catch.

    I have no idea what that means. One, that is not a definition, that is intuition at best. Second, you state that infinity is not a real number! How can infinity be open or closed or bounded or whatever if it is not part of the consideration?

    Your criticism in the first post was that: the second set cannot be closed because the boundary does not include infinity. We are telling you: infinity is not a point on the plane.

    The interval [itex][0,\infty)[/itex] is closed. There is no infinity in the boundary. We don't care because infinity is not a real number.
    Similarly, the second set you gave is closed. There is no infinity in the boundary. We don't care because infinity is not a point on the plane.
  9. Nov 27, 2011 #8


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    Then I strongly suggest you look up the definition of "closed set" in your text book or ask your teacher. You cannot do anything with a closed set if you do not kow what the definition is. In mathematics, definitions are "working definitions"- you use the exact words of the definition in proofs. It is not enough to have a vague idea of what something means- you must know the exact words of the definition.

    In metric spaces, we have a "distance" or "metric" function d(x,y) and then define the "ball centered at p with radius r" to be the set of all points whose distance from p is strictly less than r". A point, p, in a set A is called an "interior point" of A if and only if there exist a ball centered at p that is completely contained in A. A set is said to be "open" if all of its points are interior points.

    One way of defining "closed set" is that a set is closed if and only if its complement is open.

    Another way that I like is this: define "interior point" as above. Define a point to be an "exterior point" of set A if and only if it is a interior point of the complement of A. Define a point to be "boundary point" of set A if and only if it is neither an interior point nor an exterior point of A.

    Then we say that a set is "open" if it contains none of its boundary points and "closed" if it contains all of its boundary points.

    For example, in the real numbers with the "usual topology" (the metric topology where d(x,y)= |x- y|) the boundary points of the intervals (0, 1), (0, 1], [0, 1), and [0, 1] are, for each, 0 and 1. (i am using the standard "interval" notation. Each of these contains all points between 0 and 1. the "(" or ")" indicates that end point is not in the set, the "[" or "]" indicates that end point is in the set.) The first set, (0, 1), does not contain either and so is open. The last set, [0, 1] contains both and so is closed. The other two are neither open nor closed.

    But being "closed" has nothing to do with being "bounded". The sets [itex][0, \infty)[/itex] and [itex](0, \infty)[/itex] both contain all positive numbers. The first does not include 0, the second does. (Notice that both have ")" for "[itex]\infty[/itex]. Since [itex]\infty[/itex] is not a real number, it cannot be in any set of real numbers.) Both have the single boundary point, 0. The first set is open because it does not include that boundary point, the secpmd is closed because it does.

    One last example- the entire set of all real numbers has NO boundary points. It is both open and closed because it contains none of its boundary points (it can't- it has none) and it contains all of them (since it has no boundary points, none is "all".)

    Now, all of that is based on one definition of "closed" set. Since you still have not told us what definition you are using, I have no idea if any of this is relevant or not.
  10. Nov 27, 2011 #9


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    Apropos the use of definitions in mathematics, I once had an exprience that I think is worth sharing. It was my practice to start each class by asking if there were any questions about the home work. In one class it turned out the entire class was stuck on a single problem (a proof in Linear Algebra). I looked at the problem and immediately saw where the difficulty was likely to be. I wrote the entire problem on the board and underlined one word in the problem and asked the class "what does that word mean?" When I got no response, I sat at the teachers desk, put my hands on the desk and said nothing more.

    At first, of course, I got bewildered stares but, eventually, students started leafing through their texts. Finally, I think, one student actually looked the word up in the index! Once they had found the precise definition in the text, the students were able to do the problem. Again, you use the exact words of definitions in proofs.
  11. Nov 27, 2011 #10


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    the problem lies in our intuition of what "boundary" and "closed" mean in ordinary english.

    intutively, we think: a closed set should include all of its boundary.

    but what if the set itself HAS no boundary? this is certainly the case, for example, with all of R2 itself. if we try to add "the boundary points of R2" to R2, there's not anything to add.

    so we have a choice to make: we could decide to define "closed" by:

    a) S U bd(S) = S
    b) something else which excludes sets with "unbounded parts".

    the decision was made long ago, to use (a), and reserve (b) for another concept, compact (in a metric space, this means "closed and (totally) bounded").

    the advantage of using definition (a), is that it allows us to say a set S is closed if and only if its complement is open. this makes "openness" and "closedness" dual to each other, as we logically feel they should be. now look at the sets you have used in your example. it is rather clear that their complements are indeed what you would think of as "open", yes?

    (it seems rather paradoxical, but R2 is actually both open AND closed. learn to live with it).
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