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

1. 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:

If $\displaystyle U = X \cap O$ for some open subset $\displaystyle O$ of $\displaystyle \mathbb{R}$ ...

... then ...

... the subset $\displaystyle U$ of $\displaystyle X$ is open in $\displaystyle X$ ...

Help will be much appreciated ...

My thoughts so far as as follows:

Suppose $\displaystyle U = X \cap O$ for some open subset $\displaystyle O$ of $\displaystyle \mathbb{R}$ ...

Need to show $\displaystyle U$ is open in $\displaystyle X$ ... that is for every $\displaystyle p \in U$ there exists $\displaystyle \epsilon \gt 0$ such that $\displaystyle N_{ \epsilon } (p) \cap X \subset U$ ... ...

Now ... let $\displaystyle p \in U$ ...

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

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

BUT ...

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

Hope someone can help ...

Peter

2.

3. Hi Peter,

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

Originally Posted by GJA
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

5. Hi Peter,

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

Originally Posted by GJA
Hi Peter,

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

Thanks GJA ...

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

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

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

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

Is that correct?

Peter

7. Yes, this is correct. Nicely done.

Originally Posted by GJA
Yes, this is correct. Nicely done.

Thanks for all your help, GJA ...

It is much appreciated...

Peter