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Math Help - Finding an error in a proof (topology)

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    Finding an error in a proof (topology)

    I am trying to find an error in the below proof. Any advice will be highly appreciated.

    Proof: \overline{\cup A_{\alpha}} \subset \cup\bar{A_{\alpha}}

    If \{A_{\alpha}\} is a collection of sets in X and if x \in \overline{\cup A_{\alpha}} , then every neighborhood U of x intersects \cup A_{\alpha}. Then U must intersect some A_{\alpha}, so that x must belong to the closure of some A_{\alpha}. Therefore,  x \in \cup\bar{A_{\alpha}}.
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    Quote Originally Posted by aliceinwonderland View Post
    I am trying to find an error in the below proof. Any advice will be highly appreciated.

    Proof: \overline{\cup A_{\alpha}} \subset \cup\bar{A_{\alpha}}

    If \{A_{\alpha}\} is a collection of sets in X and if x \in \overline{\cup A_{\alpha}} , then every neighborhood U of x intersects \cup A_{\alpha}. Then U must intersect some A_{\alpha}, so that x must belong to the closure of some A_{\alpha}. Therefore,  x \in \cup\bar{A_{\alpha}}.
    It's true that every neighbourhood U of x must intersect some A_{\alpha}. However, different neighbourhoods may intersect different A_{\alpha}s. Your argument assumes that each U intersects a single A_{\alpha}.

    Here's an example. For n=1,2,3,..., let A_n=(1/n,1]. Then \cup A_n = (0,1], and \overline{\cup A_n} = [0,1]. But \bar{A}_n=[1/n,1], and \cup \bar{A}_n = (0,1]. So 0\in\overline{\cup A_n} but 0\notin\cup \bar{A}_n.
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