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anonymous

  • 4 years ago

Is \((0,1)\cup\{2\}\) open or closed?

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  1. anonymous
    • 4 years ago
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    Howcome? :(

  2. anonymous
    • 4 years ago
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    i would not say open because the complement is not closed

  3. anonymous
    • 4 years ago
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    This is what I did:

  4. anonymous
    • 4 years ago
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    open it is

  5. anonymous
    • 4 years ago
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    depends on your definition of open but one definition is for every element in A there is an open neighborhood contained in A that is not the case for the number 2

  6. anonymous
    • 4 years ago
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    A is open if \[\forall x \in A \exists \epsilon\text { such that } (x-\epsilon,x+\epsilon)\in A\]

  7. anonymous
    • 4 years ago
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    also clearly not closed, since it does not contain all its limit points (namely 0 and 1)

  8. anonymous
    • 4 years ago
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    Theorem: If a set \(A\) has an isolated point, then it cannot be an open set. Proof: Let \(a\in A\) be an isolated point in A. Then there exists an \(\epsilon\)-neighborhood \(V_{\epsilon}(a)\) such that \(V_{\epsilon}(a)\cap A=\{a\}\). It follows that \(A\) cannot be open since there exists no \(\epsilon\)-neighborhood \(V_{\epsilon}(a)\) such that \(V_{\epsilon}(a)\subseteq A\). \(\blacksquare\)

  9. anonymous
    • 4 years ago
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    (with \(\epsilon>0\), that is)

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