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

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Lemma 1.2.5 (ii) in Duistermaat and Kolk's "Multidimensional Real Analysis I" is limited to finite collections of open subsets of R^n, raising questions about the implications for infinite collections. The discussion highlights that the intersection of an infinite series of open cubes results in the singleton set {0}, which is not open in R^n. This illustrates a potential issue with infinite collections, as they can lead to non-open sets. The lack of explanation from the authors regarding this restriction prompts a request for clarification from the community. Understanding these limitations is crucial for grasping the concepts of continuity and open sets in higher dimensions.
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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 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
 
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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
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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