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A Confusion about open sets

  1. Mar 5, 2009 #1
    Hi all,
    I am studying Topological spaces and specially metric spaces. Suppose [0,1] U [2,3] is a metric space with some predefined metric. The the question is, Is [0,1] is open or closed or both open and closed or neither open nor closed ?
    My answer:
    A) Since every point in an open set is an interior point, then [0,1] on the metric space is not open, since one can always construct a neighbourhood arround the point 1 or 0, whose intersection with the set consists of infinite points which are not in the set [0,1]. It seems it is closed set because it contains all of its accumulation points. Now if [0,1] is closed then so is [2,3]. But the complement of the closed set in a metric space must be open and therefore [2,3] is also open. This means [2,3] is both open and closed. Tha same is true for [0,1]. Therefore it is also both open and closed.

    B) Every point in the set [0,1] is an interior point because the neighbourhood arround point 1(no matter how small it may be) consists of points which lies in the interval [0,1][tex]\bigcup[/tex][2,3]. It is because the set {x:[tex]\rho[/tex](x) [tex]\geq[/tex]1 } does not lie within the space. That means [0,1] is an open set, then same as [2,3]. but again the complement of an open set must be closed. so [2,3] must be closed. Thus [0,1] is both open and closed.

    So out of this two explanation which one is correct. If explanation two is correct then does it mean that when we talk about intersection of interior points of a point a and the set containing a, the intersection set has to lie within the inherent metric space ? Does that mean set of natural numbers {1,2,3,4,...} on R is closed while the set {2,3} is open in the space {1,2,3,4,..}.
    Please help me to sort this confusion. I am very novice in this field so please disregard if i made a very silly question.
    regards
    Titas
     
  2. jcsd
  3. Mar 5, 2009 #2

    HallsofIvy

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    Staff Emeritus
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    What closed set is [2, 3] the complement of?

    Actually, because you only said "with some predefined metric", you can't answer this question. There exist metrics with which this set is closed and metrics with which it is both open and closed. You seem to be assuming that the metric here is the standard metric on the real numbers, d(x,y)= |x-y|.

     
  4. Mar 5, 2009 #3
    Thanks HallsofIvy...
    I appreciate your advice and i will definitely go through the basic definitions. Since you questioned about the words "Some predefined metric", lets assume it is the standard metric on real numbers. For this metric what would be the answer ?

    following are some clarification
    1) [tex]\rho[/tex](x), to be read as [tex]\rho[/tex](x,0).

    2) I reframed my second explanation as:

    Every point in the set [0,1] is an interior point because the neighbourhood arround point 1(with the usual metric defined on real numbers) consists of points which lies in the interval [0,1][tex]\cup[/tex][2,3]. It is because the set {x:1<[tex]\rho[/tex](x,0)<2 } does not lie within the space. That means the neighborhood of point 1 is entirely within the defined metric space. Therefore 1 is an interior point. Same argument follows 0 is also an interior point. Therefore clearly [0,1] is an open set since it contains all the interior point. Same argument says [2,3] also open set. But again the complement of an open set within the metric space must be closed. So complement of [2,3], which is [0,1], is closed . Thus [0,1] is both open and closed.

    With the definition of usual metric on real numbers why the explanation A is not correct ? can you explain.
    You said "There exist metrics with which this set is closed and metrics with which it is both open and closed"..can you give such an example..this will enlighten me about some basics.
    Also I feel my basics is not clear enough. Can you suggest/advice some elementary books which will help me (not Rudin please).
    Thanks in advance
     
  5. Mar 6, 2009 #4

    A couple of suggestions of things to think about.

    1) [0,1] is a closed subset of the line with the usual metric. The same is true of [2,3]. But they are both open sets in the subspace topology even though the end points are still points of accumulation. Why is this?

    Generally I think you need to ponder how open and closed differ for a subset and a subset considered as a separate topological space with the subspace topology.

    2) If your metric is arbitrary it can be quite wild. The two intervals no longer have any structure that they inherit from the line. They are just sets. Try constructing metrics with the following properties:

    a) a metric on these two sets where the space is connected.
    b) another where 2 by itself is an open set but the rest of the space is connected
    c) another where the sequence 1/2 3/4 7/8 ... converges to 2.5.
     
    Last edited: Mar 6, 2009
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