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

#### Peter

##### Well-known member
MHB Site Helper
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?

Peter