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Math Help - Show A is not a closed set, but the union of A and {0} is a closed set.

  1. #1
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    Show A is not a closed set, but the union of A and {0} is a closed set.

    So basically, I have no idea how to answer this problem. I tried a few different approaches, including trying to show that the complement of A was closed. I am stuck at the very beginning as I do not know where to start with this problem. Could someone please, at the very least, outline what I am supposed to do to prove this? Thank you.

    Here's the problem:

    Show that A= \{\frac{1}{n} : n\in N\} is not a closed set, but that A \cup \{0\} is a closed set.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by seamstress View Post
    So basically, I have no idea how to answer this problem. I tried a few different approaches, including trying to show that the complement of A was closed. I am stuck at the very beginning as I do not know where to start with this problem. Could someone please, at the very least, outline what I am supposed to do to prove this? Thank you.

    Here's the problem:

    Show that A= \{\frac{1}{n} : n\in N\} is not a closed set, but that A \cup \{0\} is a closed set.
    A set is closed if it contains all of its limit points. What are the limit point(s) of A?
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    Okay, I see what you are saying.
    Right now, I have supposed c is a cluster point of A.
    I used the definition of a cluster point to say that every e-neighborhood of c contains a point x in A such that x is not equal to c
    ==> x is contained in the e-neighborhood of c= (c-e, c+e) for some e>0
    ==> |x-c| < e

    1. is this right?
    2. How would I go about showing 0 is a cluster point and that A doesn't contain it? I'm slightly confused as what to do with the e.

    thank you!

    EDIT: wow, sorry guys. just realized I made a grave mistake. Please re-read my question
    Last edited by seamstress; February 23rd 2011 at 07:12 PM.
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  4. #4
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    Please re-read my second post. I made a huge mistake while typing but just fixed it. what do you think?
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  5. #5
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    0 is a boundary point of A because every neighborhood of zero contains a point in A (namely 1/n for some n sufficiently large), and a point not in A (namely 0). So A is not closed.
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  6. #6
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    Okay, I understand that. And I also understand that 0 is the only limit point of A, but I dont know how to prove that rigorously ?
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  7. #7
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    Let U be an \epsilon neighborhood of 0. So x\in U \Leftrightarrow -\epsilon <x<\epsilon. Choose n\in \mathbb{N} such that n>\frac{1}{\epsilon}. Then \frac{1}{n}<\epsilon, and we have found an element of A in U. Thus 0 is a limit point of A.
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  8. #8
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    yes, but how would i prove that 0 is the only limit point of A?
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  9. #9
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    If x\ne \frac{1}{n} for all n, let \epsilon = \inf \{ |x-\frac{1}{n}| : n\in \mathbb{N}\}, and then argue that \epsilon > 0 and the \epsilon neighborhood of x does not contain any points of A.

    Do the case x=\frac{1}{n} separately.
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