0

Show that the set $\mathbb{R}^{n-1} \times 0$ has measure zero in $\mathbb{R}^{n}$.

Well, I resolved as follows:

We have $\mathbb{R}^{n-1} \times 0 \subset \mathbb{R}^{n}$. Let $\cup_{\lambda \in L} A_{\lambda}$ be an open cover of $\mathbb{R}^{n-1} \times 0$. So, the Theorem of Lindelof we say that $\cup A_{\lambda}$ admits an enumerable under coverage $\cup_{i=1}^{\infty} A_{\lambda_i}$. Set $$A_{\lambda_i} = ( x_{1,i} - \frac{\sqrt[n]{\epsilon}}{2\sqrt[n]{2^{i+1}}}, x_{1,i} + \frac{\sqrt[n]{\epsilon}}{2\sqrt[n]{2^{i+1}}}) \times ( x_{2,i} - \frac{\sqrt[n]{\epsilon}}{2\sqrt[n]{2^{i+1}}}, x_{2,i} + \frac{\sqrt[n]{\epsilon}}{2\sqrt[n]{2^{i+1}}}) \times \cdots \times ( x_{n-1,i} - \frac{\sqrt[n]{\epsilon}}{2\sqrt[n]{2^{i+1}}}, x_{n-1,i} + \frac{\sqrt[n]{\epsilon}}{2\sqrt[n]{2^{i+1}}}) \times ( - \frac{\sqrt[n]{\epsilon}}{2\sqrt[n]{2^{i+1}}},\frac{\sqrt[n]{\epsilon}}{2\sqrt[n]{2^{i+1}}}),$$ where $(x_{1,i},x_{2,i}, \cdots , x_{n-1,i}, 0) \in \mathbb{R}^{n-1}\times {0}.$

Being $v(A_{\lambda_i})$ the volume of $A_{\lambda_i}$. We have to $v(A_{\lambda_i}) = \left( \frac{2\sqrt[n]{\epsilon}}{2\sqrt[n]{2^{i+1}}}\right)^{n} = \frac{{\epsilon}}{{2^{i+1}}}$. So we have to $$\sum_{i=1}^{\infty} v(A_{\lambda_i}) = \epsilon \sum_{i=1}^{\infty} \frac{1}{2^{i+1}} = \frac{\epsilon}{2} < \epsilon.$$

I would like to know if this solution is correct. My biggest question is if I can do $A_{\lambda_i}$ as done above.

DANN
  • 149
  • 9
  • Is "in $\mathbf{R}^{n}$" ?What is $0$ ? Is it ${0}$? – Olórin Jul 24 '19 at 15:21
  • @DietrichBurde Unless the OP means $\varnothing$ with $0$. (I don't think it's the case though, ah ah.) – Olórin Jul 24 '19 at 15:22
  • Well, in Munkres there is only $"0"$ but I am considering it as ${ 0 }$. I don't believe it makes a difference. – DANN Jul 24 '19 at 15:24
  • Well $0$ is un number and ${0}$ is a set, so there is a difference. "In Munkres" the $0$ must denote ${0}$ for sure. – Olórin Jul 24 '19 at 15:27
  • @DANN Also, please shorten your equation so it appears correctly in the screen. :) – Olórin Jul 24 '19 at 15:28
  • I'm not sure you've shown $\mathbb R^{n-1}\times{0} \subseteq \cup_i A_{\lambda_i}$. – eyeballfrog Jul 24 '19 at 15:28
  • Hint: Take a dense sequence $x_i\in \Bbb R^{n-1}$ and then construct $A_i$ with "center" $(x_i,0)\in\Bbb R^n$. – Ryan Unger Jul 24 '19 at 15:30
  • @ujsgeyrr1f0d0d0r0h1h0j0j_juj Thank you for your observation! – DANN Jul 24 '19 at 15:33
  • @DietrichBurde I saw it and the solution is another. I would like to know if my solution is right or wrong. It's because. – DANN Jul 24 '19 at 15:35
  • @RyanUnger But why doesn't the above solution work? – DANN Jul 24 '19 at 15:37
  • The notation $a\times Y$ is pretty common and much less cumbersome than ${a}\times Y$. Admittedly, it is an abuse of notation, but where the context is clear it's much cleaner. As I recall, Munkres actually explains what he means by the notation anyway. – MPW Jul 24 '19 at 15:45
  • I don’t know if it works or not. I’m not going to read it. I’m suggesting a better and more common solution. – Ryan Unger Jul 24 '19 at 16:48
  • @DANN The Lindelof property doesn't explicitly say that the open sets of the subcovering are specifically going to be open rectangles. Much less say that they are going to be of the form that you have written. –  Jul 24 '19 at 18:28

0 Answers0