MHB Open Subsets of R^n .... D&K Lemma 1.2.5

  • Thread starter Thread starter Math Amateur
  • Start date Start date
  • Tags Tags
    Subsets
Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
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 (ii) ...

Duistermaat and Kolk"s statement and proof of Lemma 1.2.5 reads as follows: View attachment 7673My question regarding Lemma 1.2.5 is as follows:

Lemma 1.2.5 (ii) is stated and proved only for a finite collection of open subsets of $$\mathbb{R}^n$$ ... but why do we restrict the result to finite collections of open subsets ... there must be a problem with the infinite collection case ... but D&K give no explanation of why this is so ...

Can someone please explain the difficulty with the infinite collection case ...

Hope someone can help ...

Peter
 
Physics news on Phys.org
The intersection of open cubes $$C_m := \left(-\frac{1}{m}, \frac{1}{m}\right)\times\cdots \times\left(-\frac{1}{m}, \frac{1}{m}\right)\quad (m = 1,2,3,\ldots)$$ in $\Bbb R^n$ is the set containing only the origin $\bf 0$, but $\{\bf 0\}$ is not open in $\Bbb R^n$.
 
Euge said:
The intersection of open cubes $$C_m := \left(-\frac{1}{m}, \frac{1}{m}\right)\times\cdots \times\left(-\frac{1}{m}, \frac{1}{m}\right)\quad (m = 1,2,3,\ldots)$$ in $\Bbb R^n$ is the set containing only the origin $\bf 0$, but $\{\bf 0\}$ is not open in $\Bbb R^n$.
Thanks Euge,

Peter
 
Back
Top