0
$\begingroup$

I'm trying to show that the cartesian product of real numbers and the Cantor set ($C$) has (Lebesgue) measure zero in $\Bbb R^2$, I was successful at showing (I think) that $\Bbb R$ has measure zero in $\Bbb R^2$ by covering it with many "rectangles" with Volume(area) zero. Now my plan was to show that $C\times\Bbb R$ is a subset of the union of many "rectangles" and some set that contains $C$, but I can't seem to find that second set. Now, I know how to show that $C$ has measure zero in $\Bbb R$, but I can't seem to translate this into $\Bbb R^2$.

Could anybody help me? I think that it would be useful to find how to cantor set looks in R$^2$ because co-student told me that it would be the same as in R but with infinite strips going up from the small intervals, but this doesn't fit right with me.

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Hint: First show that $S_n = (C \times \Bbb R) \cap ([-n,n] \times [-n,n])$ has measure $0$ for any $n \in \Bbb N$. Using the fact that $C \times \Bbb R = \bigcup_{n \in \Bbb N} S_n$, conclude that $C \times \Bbb R$ has measure $0$.

$\endgroup$
3
  • $\begingroup$ I have difficulty showing that $\textit{S$_n$}$ has measure $0$. Would you be so kind and provide anothet tip? $\endgroup$
    – Nitaa a
    Nov 4, 2020 at 18:46
  • 1
    $\begingroup$ @Nitaaa You can cover the cantor set with intervals whose total length is arbitrarily small. use these intervals to make rectangles that cover $S_n$ and have arbitrarily small total area. $\endgroup$
    – Ben Grossmann
    Nov 4, 2020 at 18:48
  • $\begingroup$ Ah yes, I get it now, thank you very much. $\endgroup$
    – Nitaa a
    Nov 4, 2020 at 18:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .