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Math Help - Proofs -_____-

  1. #1
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    Proofs -_____-

    Can anyone help me prove the following?
    Suppose S is a set of numbers which contain exactly one element, then S is closed.

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    Senior Member jakncoke's Avatar
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    Re: Proofs -_____-

    I'm not sure i understand the question. Do you mean like S = { {1}, {2}, {3} } etc... ? or S = {1}.
    Last edited by jakncoke; November 18th 2012 at 11:15 PM.
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    Re: Proofs -_____-

    like S = {1}.
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    Senior Member jakncoke's Avatar
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    Re: Proofs -_____-

    Quote Originally Posted by duckyone View Post
    like S = {1}.
    If S contains only one element then it is the empty set no? Since the empty set is a subset of every set.
    if the element is the empty set. Then vacously  {\phi} + {\phi} = {\phi} \in S

    but even if they mean S = {x}, you merely say that  {x} \cup {x} = {x} \in S and  {x} \cup {\phi} = {x} \in S
    Last edited by jakncoke; November 18th 2012 at 11:21 PM.
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    Re: Proofs -_____-

    Quote Originally Posted by jakncoke View Post
    If S contains only one element then it is the empty set no? Since the empty set is a subset of every set.
    if the element is the empty set. Then vacously  {\phi} + {\phi} = {\phi} \in S

    but even if they mean S = {x}, you merely say that  {x} \cup {x} = {x} \in S and  {x} \cup {\phi} = {x} \in S
    I can't make heads or tails out of what you are saying! "If S contains only one element then it is the empty set no?" No! The empty set has no elements! You may be confusing "element" with "subset". And \{x\}\cup \{x\}= \{x\} \in S makes no sense because your S does not contain {x}, it contains only x.

    Further, none of that has anything to do with whether such a set is closed or not.

    duckyone, this is NOT in general true. It is true in a metric space so I will assume that is what you mean. There are a variety of ways, even in a metric space, of defining "closed" set and I don't know which you are using. One I like is this: we say that a point, p, is an "interior point" of set A if and only if there exist \delta> 0 such that the neighborhood centered on p with radius \delta is a subset of A. We say that p is an "exterior" point of set A if and only if it is an interior point of the complement of A. Finally, we say that p is a "boundary" point of A if and only if it is neither an interior point or exterior point of A.

    We then say that a set is "open" if and only if it contains none of its boundary points and that a set is "closed" if and only if it contains all of its boundary points. Here, Taking A= {x}, if y\ne x, there is some fixed distance between x and y, d(x,y). Taking \delta= d(x,y)/2, it is clear that the neighborhood of y, with radius \delta, does NOT contain x so that y is an exterior point. That is, every y\ne x is an exterior point of A. That means that the set of boundary points of A is either empty or {x} itself. In either case, A contains its set of boundary points and so is closed.

    If your course did not introduce "boundary points" other commonly used definition is that a set is closed if and only if its complement is open. And, of course, a set is open if and only if every point is an interior point. This case, the complement of A= {x} is the set all points other than x. And every such point is, as shown above, an interior point of the complement of A.
    Last edited by HallsofIvy; November 19th 2012 at 08:23 AM.
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    Re: Proofs -_____-

    let us suppose you are talking about a set of REAL numbers.

    DEFINITION: a subset S of the real numbers R is open if and only if it is a union of (perhaps infinitely many) open intervals (a,b) = {x in R: a < x < b}.

    DEFINITION: a subset S of R is closed if and only if R - S is open.

    THEOREM: the set S = {x0} is closed.

    PROOF:

    R - S = (-∞,x0) U (x0,∞) = A U B.

    consider the open intervals of the form: Ak = (x0-k,x0), where k is a positive integer.

    then:

    \bigcup_{k \in \mathbb{Z}^+} A_k = A.

    similarly, for Bk = (x0,x0+k) we have:

    \bigcup_{k \in \mathbb{Z}^+} B_k = B.

    so A and B are open, so A U B is also open (being an even BIGGER union of infinitely many open intervals), so R - S is open, so S is closed.

    ***************

    note that HallsOfIvy's post is much more general, and subsumes mine (the real line is a special case of a metric space: for the interval (a,b) we may use the epsilon: ε = (b-a)/2 in which case-

    (a,b) = ((a+b)/2 - ε,(a+b)/2 + ε) = {x in R: d(x,(a+b)/2) < ε}

    the metric function d in this case is d(x,y) = |y - x|).

    ***************

    in the most general formulation possible (point-set topology), it is NOT true that singleton sets need be closed. for example, we can declare the only subsets of a set X that are open are the null set and X itself. for {x} to be closed (where x is some element of X), we would need X - {x} to be open. if X contains more than just the single point {x},

    then X - {x} is neither nor X itself, and as such cannot be open. this unhappy state of affairs CAN happen (such a topology is called the "indiscrete topology" and can be imagined like so: X is a "tightly bound blob" whose elements are all "so tightly glued together we can't separate them").
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