anonymous
  • anonymous
Is \((0,1)\cup\{2\}\) open or closed?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Howcome? :(
anonymous
  • anonymous
i would not say open because the complement is not closed
anonymous
  • anonymous
This is what I did:

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anonymous
  • anonymous
open it is
anonymous
  • anonymous
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
anonymous
  • anonymous
A is open if \[\forall x \in A \exists \epsilon\text { such that } (x-\epsilon,x+\epsilon)\in A\]
anonymous
  • anonymous
also clearly not closed, since it does not contain all its limit points (namely 0 and 1)
anonymous
  • anonymous
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 \(V_{\epsilon}(a)\subseteq A\). \(\blacksquare\)
anonymous
  • anonymous
(with \(\epsilon>0\), that is)

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