Skip to main content
MathOverflow

Return to Question

a fuller sentence
Source Link
Wlod AA
Wlod AA
  • 4.8k
  • 17
  • 23

Let me quote en.wikipedia about the original Nikodym's example (the definitions above I've written myself just for MO):

a Nikodym set is a subset of the unit square in $\ \mathbb R ^2\ $ with the complement of Lebesgue measure zero, such that, given any point in the set, there is a straight line that only intersects the set at that point. The existence of a Nikodym set was first proved by Otto Nikodym in 1927.

End of quote. See https://en.wikipedia.org/wiki/Nikodym_set.

In the above example, we can talk about the open square (or else, we may remove the border points of the square from the set and the things would still work). The $\ L\ $ would be the intersections of the straight lines with the open square. This would define a little bit nicer Nikodym set.

I am surprised that the original Nikodym set was provided in $\ (0;1)^2\ $ instead of $\ \mathbb R^2,\ $hence

Question 1: Let $\ L\ $ be the set of all straight lines in $\ \mathbb R^2\ $ Does there exist a respective Nikodym set, i.e. $\ A\subseteq \mathbb R^2\ $ such that the complement of $\ A\ $ has measure $0\,$ and $$ \forall_{x\in A}\exists_{\ell\in L}\quad \ell\cap A=\{x\} $$ ?

Question 2: Does there exist $\ A\ $ as in  Question 1  and such that $\ A = \{2\cdot x: x\in A\}\ $ ?

I'll stop now while additional natural questions come to one's mind; some of them after reading the whole quoted en.wikipedia article.

Let me quote en.wikipedia about the original Nikodym's example (the definitions above I've written myself just for MO):

a Nikodym set is a subset of the unit square in $\ \mathbb R ^2\ $ with the complement of Lebesgue measure zero, such that, given any point in the set, there is a straight line that only intersects the set at that point. The existence of a Nikodym set was first proved by Otto Nikodym in 1927.

End of quote. See https://en.wikipedia.org/wiki/Nikodym_set.

In the above example, we can talk about the open square (or else, we may remove the border from the set and the things would still work). The $\ L\ $ would be the intersections of the straight lines with the open square. This would define a little bit nicer Nikodym set.

I am surprised that the original Nikodym set was provided in $\ (0;1)^2\ $ instead of $\ \mathbb R^2,\ $hence

Question 1: Let $\ L\ $ be the set of all straight lines in $\ \mathbb R^2\ $ Does there exist a respective Nikodym set, i.e. $\ A\subseteq \mathbb R^2\ $ such that the complement of $\ A\ $ has measure $0\,$ and $$ \forall_{x\in A}\exists_{\ell\in L}\quad \ell\cap A=\{x\} $$ ?

Question 2: Does there exist $\ A\ $ as in  Question 1  and such that $\ A = \{2\cdot x: x\in A\}\ $ ?

I'll stop now while additional natural questions come to one's mind; some of them after reading the whole quoted en.wikipedia article.

Let me quote en.wikipedia about the original Nikodym's example (the definitions above I've written myself just for MO):

a Nikodym set is a subset of the unit square in $\ \mathbb R ^2\ $ with the complement of Lebesgue measure zero, such that, given any point in the set, there is a straight line that only intersects the set at that point. The existence of a Nikodym set was first proved by Otto Nikodym in 1927.

End of quote. See https://en.wikipedia.org/wiki/Nikodym_set.

In the above example, we can talk about the open square (or else, we may remove the border points of the square from the set and the things would still work). The $\ L\ $ would be the intersections of the straight lines with the open square. This would define a little bit nicer Nikodym set.

I am surprised that the original Nikodym set was provided in $\ (0;1)^2\ $ instead of $\ \mathbb R^2,\ $hence

Question 1: Let $\ L\ $ be the set of all straight lines in $\ \mathbb R^2\ $ Does there exist a respective Nikodym set, i.e. $\ A\subseteq \mathbb R^2\ $ such that the complement of $\ A\ $ has measure $0\,$ and $$ \forall_{x\in A}\exists_{\ell\in L}\quad \ell\cap A=\{x\} $$ ?

Question 2: Does there exist $\ A\ $ as in  Question 1  and such that $\ A = \{2\cdot x: x\in A\}\ $ ?

I'll stop now while additional natural questions come to one's mind; some of them after reading the whole quoted en.wikipedia article.

Source Link
Wlod AA
Wlod AA
  • 4.8k
  • 17
  • 23

Generalized Nikodym sets

Let me quote en.wikipedia about the original Nikodym's example (the definitions above I've written myself just for MO):

a Nikodym set is a subset of the unit square in $\ \mathbb R ^2\ $ with the complement of Lebesgue measure zero, such that, given any point in the set, there is a straight line that only intersects the set at that point. The existence of a Nikodym set was first proved by Otto Nikodym in 1927.

End of quote. See https://en.wikipedia.org/wiki/Nikodym_set.

In the above example, we can talk about the open square (or else, we may remove the border from the set and the things would still work). The $\ L\ $ would be the intersections of the straight lines with the open square. This would define a little bit nicer Nikodym set.

I am surprised that the original Nikodym set was provided in $\ (0;1)^2\ $ instead of $\ \mathbb R^2,\ $hence

Question 1: Let $\ L\ $ be the set of all straight lines in $\ \mathbb R^2\ $ Does there exist a respective Nikodym set, i.e. $\ A\subseteq \mathbb R^2\ $ such that the complement of $\ A\ $ has measure $0\,$ and $$ \forall_{x\in A}\exists_{\ell\in L}\quad \ell\cap A=\{x\} $$ ?

Question 2: Does there exist $\ A\ $ as in  Question 1  and such that $\ A = \{2\cdot x: x\in A\}\ $ ?

I'll stop now while additional natural questions come to one's mind; some of them after reading the whole quoted en.wikipedia article.