
MHB Master
#1
January 10th, 2020,
02:46
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:
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

January 10th, 2020 02:46
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