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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:

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

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