Relatively open sets in Stoll's Theorem 3.1.16

[Math Amateur]

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

I need 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 9514Can someone please help me to demonstrate a formal and rigorous proof of the following:If $$U = X \cap O$$ for some open subset $$O$$ of $$\mathbb{R}$$ ...

... then ...

... the subset $$U$$ of $$X$$ is open in $$X$$ ...
Help will be much appreciated ...
My thoughts so far as as follows:

Suppose $$U = X \cap O$$ for some open subset $$O$$ of $$\mathbb{R}$$ ...Need to show $$U$$ is open in $$X$$ ... that is for every $$p \in U$$ there exists $$\epsilon \gt 0$$ such that $$N_{ \epsilon } (p) \cap X \subset U$$ ... ... Now ... let $$p \in U$$ ...

then $$p \in O$$ ...

Therefore there exists $$\epsilon \gt 0$$ such that $$N_{ \epsilon } (p) \subset O$$ ... since $$O$$ is open ...BUT ...

... how do I proceed from here ... ?

Hope someone can help ...

Peter

 

[GJA]

Hi Peter,

Everything looks good so far. From here, what can be said about $N_{\epsilon}(p)\cap X$?

 

[Math Amateur]

Hi Peter,

Everything looks good so far. From here, what can be said about $N_{\epsilon}(p)\cap X$?

Hi GJA ...

Still perplexed ... can you help further...

Peter

 

[GJA]

Hi Peter,

Think about trying to use $N_{\epsilon}(p)\subset O$ and use that fact to get a set "inequality" for $N_{\epsilon}(p)\cap X$.

 

[Math Amateur]

Hi Peter,

Think about trying to use $N_{\epsilon}(p)\subset O$ and use that fact to get a set "inequality" for $N_{\epsilon}(p)\cap X$.

Thanks GJA ...

I think the argument you're suggesting is as follows:

We have $N_{\epsilon}(p)\subset O$

So therefore $$N_{\epsilon}(p) \cap X \subset O \cap X$$ ...

... that is $$N_{\epsilon}(p) \cap X \subset U$$ ... as required ...Is that correct?

Peter

 

[GJA]

Yes, this is correct. Nicely done.

 

[Math Amateur]

Yes, this is correct. Nicely done.

Thanks for all your help, GJA ...

It is much appreciated...

Peter