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    I am reading Manfred Stoll's book: Introduction to Real Analysis.

    I need further help with Stoll's proof of Theorem 3.1.16

    Stoll's statement of Theorem 3.1.16 and its proof reads as follows:







    Can someone please help me to demonstrate a formal and rigorous proof of the following:


    If the subset $ \displaystyle U$ of $ \displaystyle X$ is open in $ \displaystyle X$ ...

    ... then ...

    $ \displaystyle U = X \cap O$ for some open subset $ \displaystyle O$ of $ \displaystyle \mathbb{R}$ ...



    Help will be much appreciated ...



    My attempt at a proof is as follows:


    Assume that $ \displaystyle U$ is open in $ \displaystyle X$.

    Then for every $ \displaystyle p \in U \ \exists \ \epsilon \gt 0$ such that $ \displaystyle N_{ \epsilon_p } (p) \cap X \subset U$ ...

    Then the set $ \displaystyle \cup \{ N_{ \epsilon_p } (p) \ : \ p \in U \}$ is open ... since it is the union of open sets ...

    Now we claim that $ \displaystyle \cup \{ N_{ \epsilon_p } (p) \ : \ p \in U \} \cap X = U$ as required ...


    Proof that $ \displaystyle \cup \{ N_{ \epsilon_p } (p) \ : \ p \in U \} \cap X = U$ proceeds as follows:


    Let $ \displaystyle x \in \cup \{ N_{ \epsilon_p } (p) \ : \ p \in U \} \cap X$

    Then $ \displaystyle x \in U$ and $ \displaystyle x \in X$ ... so obviously $ \displaystyle x \in U$ ... ...



    Let $ \displaystyle x \in U$

    Then $ \displaystyle x \in N_{ \epsilon_x } (x) \cap X$

    Therefore $ \displaystyle x \in \cup \{ N_{ \epsilon_p } (p) \ : \ p \in U \} \cap X$




    Is the above proof correct?

    Peter

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