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# Thread: Relatively Open Sets ... Another Queston Regarding Stoll, Theorem 3.1.16 (a) ...

1. 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:

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?

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