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

[Math Amateur]

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:
View attachment 9515
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

 

[jvicens]

Your proof looks correct to me. Let me explain the steps in more detail to make it clearer.

First, we know that U is open in X, so for every point p in U, there exists an open ball N_{\epsilon_p}(p) that is contained in U. This is the definition of an open set.

Next, we take the union of all these open balls. This set is also open because it is the union of open sets. This is a key property of open sets.

Now we claim that the intersection of this set with X is equal to U. To prove this, we need to show that any point x in U is also in the intersection, and any point in the intersection is also in U.

For the first part, if x is in U, then by definition, there exists an open ball N_{\epsilon_x}(x) that is contained in U. But since x is also in X, this ball is also contained in X. Therefore, x is in the intersection.

For the second part, if x is in the intersection, then it is in an open ball N_{\epsilon_p}(p) for some point p in U. But since the intersection is with X, this ball must also be contained in X. Therefore, x is in U.

This completes the proof that U is equal to the intersection of the union of open balls with X. This intersection is also an open set, since it is the intersection of open sets.

I hope this helps clarify the proof for you. Let me know if you have any further questions.