MHB Another question on Open and Closed in V .... D&K Proposition 1.2.17 .... ....

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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 another aspect of the proof of Proposition 1.2.17 ... ...

Duistermaat and Kolk's Proposition 1.2.17 and the preceding definition (regarding open and closed sets in a set V) read as follows:
View attachment 7735
View attachment 7736D&K write that the proof of (i) is immediate from the definitions ... but I have been unable to formulate a rigorous proof of (i) ... could someone please demonstrate a rigorous proof of (i) ... ...Hope that someone can help ...

Peter========================================================================================
D&K's definitions and early results on open and closed sets may be helpful to MHB members reading and following the above post ... so I am providing the same ... as follows:View attachment 7737
View attachment 7738Hope that helps ... ...

Peter
 
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Hi, Peter.

Suppose $A$ is open in $V$. By Definition 1.2.16 there is an open subset, say U, of $\mathbb{R}^{n}$ such that $A=V\cap U.$ Since both $U$ and $V$ are open in $\mathbb{R}^{n}$ and finite intersections of open sets are open, $A$ is open in $\mathbb{R}^{n}.$

If $A$ is open in $\mathbb{R}^{n}$, then it is open in $V$ because $A=V\cap A$ (since $A\subseteq V$).
 
GJA said:
Hi, Peter.

Suppose $A$ is open in $V$. By Definition 1.2.16 there is an open subset, say U, of $\mathbb{R}^{n}$ such that $A=V\cap U.$ Since both $U$ and $V$ are open in $\mathbb{R}^{n}$ and finite intersections of open sets are open, $A$ is open in $\mathbb{R}^{n}.$

If $A$ is open in $\mathbb{R}^{n}$, then it is open in $V$ because $A=V\cap A$ (since $A\subseteq V$).
THanks GJA ... appreciate the help ...

Peter
 
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