Open Sets in R^n .... Duistermaat and Kolk, Lemma 1.2.5 ....

[Math Amateur]

I am reading "Multidimensional Real Analysis I: Differentiation by J. J. Duistermaat and J. A. C. Kolk ...

I am focused on Chapter 1: Continuity ... ...

I need help with an aspect of Lemma 1.2.5 ...

Duistermaat and Kolk"s Lemma 1.2.5 reads as follows:View attachment 9014In the above proof by Duistermaat and Kolk we read the following:

" ... ... Assertion (i) follows from Definition 1.2.2 ... ..."I have tried to demonstrate a rigorous proof of Assertion (i) but have not been happy it is fully rigorous ...Can someone please demonstrate a fully rigorous proof of Assertion (i) ...


Help will be appreciated ...

Peter ========================================================================================It may help readers of the above post to have access to the start of Section 1.2: Open and Closed Sets ... which includes Definition 1.2.2 referred to above ... so I am providing aces to that text ... as follows ... View attachment 9015
View attachment 9016
Hope that helps ... ...

Peter

 

[Opalg]

Suppose that $U = \bigcup_{s\in S}U_s$ is a union of open sets $U_s$. If $x\in U$ then by definition $x$ must belong to one of the sets $U_s$. Since $U_s$ is open, there exists $\delta>0$ with $B(x,\delta)\subseteq U_s$. Then $x\in B(x,\delta)\subseteq U_s \subseteq U$. Thus every point $x$ in $U$ is in the interior of $U$. So $U$ is open.

 

[Math Amateur]

Suppose that $U = \bigcup_{s\in S}U_s$ is a union of open sets $U_s$. If $x\in U$ then by definition $x$ must belong to one of the sets $U_s$. Since $U_s$ is open, there exists $\delta>0$ with $B(x,\delta)\subseteq U_s$. Then $x\in B(x,\delta)\subseteq U_s \subseteq U$. Thus every point $x$ in $U$ is in the interior of $U$. So $U$ is open.

Thanks for your help, Opalg ...

Peter