Can planar set contain even many vertices of every unit equilateral triangle?Move one element of finite set out from A in planeOmega_1 unions of null sets: Martin's AxiomFollow up question on union of disjoint Vitali sets…Are Vitali-type nonmeasurable sets determinate?Are faces of a compact, convex body “opposed” iff their extreme points are pairwise “opposed”?Is it consistent with ZFC that there is a translation-invariant extension of Lebesgue measure that assigns nonzero measure to some set of measure less than c?For Every Measure Zero Set $E$ There Exists a Positive Measure with Lower Lebesgue Density 0 and Upper Lebesgue Density 1A question regarding a common critique of Freiling's Axiom of SymmetryIteration of random realsDoes there exist a Lebesgue nonmeasurable set $E$ in $mathbbR$ satisfies that $Ecap A$ is a Borel null set for every Borel null set $A$?
Can planar set contain even many vertices of every unit equilateral triangle?
Move one element of finite set out from A in planeOmega_1 unions of null sets: Martin's AxiomFollow up question on union of disjoint Vitali sets…Are Vitali-type nonmeasurable sets determinate?Are faces of a compact, convex body “opposed” iff their extreme points are pairwise “opposed”?Is it consistent with ZFC that there is a translation-invariant extension of Lebesgue measure that assigns nonzero measure to some set of measure less than c?For Every Measure Zero Set $E$ There Exists a Positive Measure with Lower Lebesgue Density 0 and Upper Lebesgue Density 1A question regarding a common critique of Freiling's Axiom of SymmetryIteration of random realsDoes there exist a Lebesgue nonmeasurable set $E$ in $mathbbR$ satisfies that $Ecap A$ is a Borel null set for every Borel null set $A$?
$begingroup$
Is there a nonempty planar set that contains $0$ or $2$ vertices from each unit equilateral triangle?
I know that such a set cannot be measurable. In fact, my motivation is to extend a Falconer-Croft proof that works for measurable sets, see the details here. In more general, to make their proof work for other sets besides equilateral triangles, one can ask the following.
Suppose we are given two sets, $S$ and $A$ in the plane, such that $S$ is finite, with a special point, $s_0$, while neither $A$ nor its complement is a null-set, i.e., the outer Lebesgue measure of $A$ and $A^c=mathbb R^2setminus A$ are both non-zero. Can we find two congruent copies of $S$, $f_1(S)$ and $f_2(S)$, such that $f_1^-1(f_1(S)cap A)Delta f_2^-1(f_2(S)cap A)=s_0$, i.e., $s_0$ is the only element of $S$ that goes in to/out of $A$ when we go from $S_1$ to $S_2$?
set-theory real-analysis mg.metric-geometry polymath16
asked 9 hours ago
domotorpdomotorp
9,95433 silver badges92 bronze badges
$endgroup$
add a comment |
$begingroup$
Is there a nonempty planar set that contains $0$ or $2$ vertices from each unit equilateral triangle?
I know that such a set cannot be measurable. In fact, my motivation is to extend a Falconer-Croft proof that works for measurable sets, see the details here. In more general, to make their proof work for other sets besides equilateral triangles, one can ask the following.
Suppose we are given two sets, $S$ and $A$ in the plane, such that $S$ is finite, with a special point, $s_0$, while neither $A$ nor its complement is a null-set, i.e., the outer Lebesgue measure of $A$ and $A^c=mathbb R^2setminus A$ are both non-zero. Can we find two congruent copies of $S$, $f_1(S)$ and $f_2(S)$, such that $f_1^-1(f_1(S)cap A)Delta f_2^-1(f_2(S)cap A)=s_0$, i.e., $s_0$ is the only element of $S$ that goes in to/out of $A$ when we go from $S_1$ to $S_2$?
set-theory real-analysis mg.metric-geometry polymath16
asked 9 hours ago
domotorpdomotorp
9,95433 silver badges92 bronze badges
$endgroup$
$begingroup$
If I understand this, you ask for a set with certain properties, one of which is that for every point in the set, the set has "half the points" of the unit circle with the first point as its center. I suspect the answer is that the empty set is such a set, and that there is no other. Gerhard "Of Course Suspicion Isn't Proof" Paseman, 2019.08.17.
$endgroup$
– Gerhard Paseman
9 hours ago
add a comment |
$begingroup$
Is there a nonempty planar set that contains $0$ or $2$ vertices from each unit equilateral triangle?
I know that such a set cannot be measurable. In fact, my motivation is to extend a Falconer-Croft proof that works for measurable sets, see the details here. In more general, to make their proof work for other sets besides equilateral triangles, one can ask the following.
Suppose we are given two sets, $S$ and $A$ in the plane, such that $S$ is finite, with a special point, $s_0$, while neither $A$ nor its complement is a null-set, i.e., the outer Lebesgue measure of $A$ and $A^c=mathbb R^2setminus A$ are both non-zero. Can we find two congruent copies of $S$, $f_1(S)$ and $f_2(S)$, such that $f_1^-1(f_1(S)cap A)Delta f_2^-1(f_2(S)cap A)=s_0$, i.e., $s_0$ is the only element of $S$ that goes in to/out of $A$ when we go from $S_1$ to $S_2$?
set-theory real-analysis mg.metric-geometry polymath16
asked 9 hours ago
domotorpdomotorp
9,95433 silver badges92 bronze badges
$endgroup$
Is there a nonempty planar set that contains $0$ or $2$ vertices from each unit equilateral triangle?
I know that such a set cannot be measurable. In fact, my motivation is to extend a Falconer-Croft proof that works for measurable sets, see the details here. In more general, to make their proof work for other sets besides equilateral triangles, one can ask the following.
Suppose we are given two sets, $S$ and $A$ in the plane, such that $S$ is finite, with a special point, $s_0$, while neither $A$ nor its complement is a null-set, i.e., the outer Lebesgue measure of $A$ and $A^c=mathbb R^2setminus A$ are both non-zero. Can we find two congruent copies of $S$, $f_1(S)$ and $f_2(S)$, such that $f_1^-1(f_1(S)cap A)Delta f_2^-1(f_2(S)cap A)=s_0$, i.e., $s_0$ is the only element of $S$ that goes in to/out of $A$ when we go from $S_1$ to $S_2$?
set-theory real-analysis mg.metric-geometry polymath16
set-theory real-analysis mg.metric-geometry polymath16
asked 9 hours ago
domotorpdomotorp
9,95433 silver badges92 bronze badges
asked 9 hours ago
domotorpdomotorp
9,95433 silver badges92 bronze badges
asked 9 hours ago
domotorpdomotorp
9,95433 silver badges92 bronze badges
asked 9 hours ago
domotorpdomotorp
9,95433 silver badges92 bronze badges
asked 9 hours ago
domotorpdomotorp
9,95433 silver badges92 bronze badges
9,95433 silver badges92 bronze badges
$begingroup$
If I understand this, you ask for a set with certain properties, one of which is that for every point in the set, the set has "half the points" of the unit circle with the first point as its center. I suspect the answer is that the empty set is such a set, and that there is no other. Gerhard "Of Course Suspicion Isn't Proof" Paseman, 2019.08.17.
$endgroup$
– Gerhard Paseman
9 hours ago
add a comment |
$begingroup$
If I understand this, you ask for a set with certain properties, one of which is that for every point in the set, the set has "half the points" of the unit circle with the first point as its center. I suspect the answer is that the empty set is such a set, and that there is no other. Gerhard "Of Course Suspicion Isn't Proof" Paseman, 2019.08.17.
$endgroup$
– Gerhard Paseman
9 hours ago
$begingroup$
If I understand this, you ask for a set with certain properties, one of which is that for every point in the set, the set has "half the points" of the unit circle with the first point as its center. I suspect the answer is that the empty set is such a set, and that there is no other. Gerhard "Of Course Suspicion Isn't Proof" Paseman, 2019.08.17.
$endgroup$
– Gerhard Paseman
9 hours ago
$begingroup$
If I understand this, you ask for a set with certain properties, one of which is that for every point in the set, the set has "half the points" of the unit circle with the first point as its center. I suspect the answer is that the empty set is such a set, and that there is no other. Gerhard "Of Course Suspicion Isn't Proof" Paseman, 2019.08.17.
$endgroup$
– Gerhard Paseman
9 hours ago
add a comment |
1 Answer
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$begingroup$
I am probably misunderstanding something. Let me suppose a non empty planar set with the property, and that point A is in the set. Pick a unit equilateral triangle having A as a vertex, and also B and C.
Then exactly one of B and C is in the set. If we pick unit equilateral triangle BCD, then D (short for Different from A) must be in the set with A. So if A is in the set, then every point with distance r (where r^2 is 3) from A is in the set.
The contradiction comes from picking arbitrary point X and stepping to it from A using steps of length r, which I leave to you.
Afterthought For your more general question, I would consider studying the following scheme, which may be in the literature: Given S, consider A-colorings of S. These are labellings of S depending on your allowed coloring scheme. For the general problem above, let red correspond to "out of A", green to "not out of A", and you color according to how you transform S to a domain containing A. For the problem above, you color a triplet corresponding to the appearance of vertices of a unit equilateral triangle inside or outside the set A.
Now a challenging problem is to determine what A-colorings are interesting/feasible, when you put restrictions on the set of transforms of S. You want to find two colorings which differ precisely at the point $s_0$. You may be able to prove existence by cardinality considerations on the set of all allowed A-colorings.
Gerhard "Please Show Me What's Wrong" Paseman, 2019.08.18.
answered 9 hours ago
Gerhard PasemanGerhard Paseman
9,7362 gold badges22 silver badges47 bronze badges
$endgroup$
$begingroup$
Well, I better undelete my original question then!
$endgroup$
– domotorp
8 hours ago
add a comment |
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$begingroup$
I am probably misunderstanding something. Let me suppose a non empty planar set with the property, and that point A is in the set. Pick a unit equilateral triangle having A as a vertex, and also B and C.
Then exactly one of B and C is in the set. If we pick unit equilateral triangle BCD, then D (short for Different from A) must be in the set with A. So if A is in the set, then every point with distance r (where r^2 is 3) from A is in the set.
The contradiction comes from picking arbitrary point X and stepping to it from A using steps of length r, which I leave to you.
Afterthought For your more general question, I would consider studying the following scheme, which may be in the literature: Given S, consider A-colorings of S. These are labellings of S depending on your allowed coloring scheme. For the general problem above, let red correspond to "out of A", green to "not out of A", and you color according to how you transform S to a domain containing A. For the problem above, you color a triplet corresponding to the appearance of vertices of a unit equilateral triangle inside or outside the set A.
Now a challenging problem is to determine what A-colorings are interesting/feasible, when you put restrictions on the set of transforms of S. You want to find two colorings which differ precisely at the point $s_0$. You may be able to prove existence by cardinality considerations on the set of all allowed A-colorings.
Gerhard "Please Show Me What's Wrong" Paseman, 2019.08.18.
answered 9 hours ago
Gerhard PasemanGerhard Paseman
9,7362 gold badges22 silver badges47 bronze badges
$endgroup$
$begingroup$
Well, I better undelete my original question then!
$endgroup$
– domotorp
8 hours ago
add a comment |
$begingroup$
I am probably misunderstanding something. Let me suppose a non empty planar set with the property, and that point A is in the set. Pick a unit equilateral triangle having A as a vertex, and also B and C.
Then exactly one of B and C is in the set. If we pick unit equilateral triangle BCD, then D (short for Different from A) must be in the set with A. So if A is in the set, then every point with distance r (where r^2 is 3) from A is in the set.
The contradiction comes from picking arbitrary point X and stepping to it from A using steps of length r, which I leave to you.
Afterthought For your more general question, I would consider studying the following scheme, which may be in the literature: Given S, consider A-colorings of S. These are labellings of S depending on your allowed coloring scheme. For the general problem above, let red correspond to "out of A", green to "not out of A", and you color according to how you transform S to a domain containing A. For the problem above, you color a triplet corresponding to the appearance of vertices of a unit equilateral triangle inside or outside the set A.
Now a challenging problem is to determine what A-colorings are interesting/feasible, when you put restrictions on the set of transforms of S. You want to find two colorings which differ precisely at the point $s_0$. You may be able to prove existence by cardinality considerations on the set of all allowed A-colorings.
Gerhard "Please Show Me What's Wrong" Paseman, 2019.08.18.
answered 9 hours ago
Gerhard PasemanGerhard Paseman
9,7362 gold badges22 silver badges47 bronze badges
$endgroup$
$begingroup$
Well, I better undelete my original question then!
$endgroup$
– domotorp
8 hours ago
add a comment |
$begingroup$
I am probably misunderstanding something. Let me suppose a non empty planar set with the property, and that point A is in the set. Pick a unit equilateral triangle having A as a vertex, and also B and C.
Then exactly one of B and C is in the set. If we pick unit equilateral triangle BCD, then D (short for Different from A) must be in the set with A. So if A is in the set, then every point with distance r (where r^2 is 3) from A is in the set.
The contradiction comes from picking arbitrary point X and stepping to it from A using steps of length r, which I leave to you.
Afterthought For your more general question, I would consider studying the following scheme, which may be in the literature: Given S, consider A-colorings of S. These are labellings of S depending on your allowed coloring scheme. For the general problem above, let red correspond to "out of A", green to "not out of A", and you color according to how you transform S to a domain containing A. For the problem above, you color a triplet corresponding to the appearance of vertices of a unit equilateral triangle inside or outside the set A.
Now a challenging problem is to determine what A-colorings are interesting/feasible, when you put restrictions on the set of transforms of S. You want to find two colorings which differ precisely at the point $s_0$. You may be able to prove existence by cardinality considerations on the set of all allowed A-colorings.
Gerhard "Please Show Me What's Wrong" Paseman, 2019.08.18.
answered 9 hours ago
Gerhard PasemanGerhard Paseman
9,7362 gold badges22 silver badges47 bronze badges
$endgroup$
I am probably misunderstanding something. Let me suppose a non empty planar set with the property, and that point A is in the set. Pick a unit equilateral triangle having A as a vertex, and also B and C.
Then exactly one of B and C is in the set. If we pick unit equilateral triangle BCD, then D (short for Different from A) must be in the set with A. So if A is in the set, then every point with distance r (where r^2 is 3) from A is in the set.
The contradiction comes from picking arbitrary point X and stepping to it from A using steps of length r, which I leave to you.
Afterthought For your more general question, I would consider studying the following scheme, which may be in the literature: Given S, consider A-colorings of S. These are labellings of S depending on your allowed coloring scheme. For the general problem above, let red correspond to "out of A", green to "not out of A", and you color according to how you transform S to a domain containing A. For the problem above, you color a triplet corresponding to the appearance of vertices of a unit equilateral triangle inside or outside the set A.
Now a challenging problem is to determine what A-colorings are interesting/feasible, when you put restrictions on the set of transforms of S. You want to find two colorings which differ precisely at the point $s_0$. You may be able to prove existence by cardinality considerations on the set of all allowed A-colorings.
Gerhard "Please Show Me What's Wrong" Paseman, 2019.08.18.
answered 9 hours ago
Gerhard PasemanGerhard Paseman
9,7362 gold badges22 silver badges47 bronze badges
edited 8 hours ago
answered 9 hours ago
Gerhard PasemanGerhard Paseman
9,7362 gold badges22 silver badges47 bronze badges
answered 9 hours ago
Gerhard PasemanGerhard Paseman
9,7362 gold badges22 silver badges47 bronze badges
answered 9 hours ago
Gerhard PasemanGerhard Paseman
9,7362 gold badges22 silver badges47 bronze badges
9,7362 gold badges22 silver badges47 bronze badges
$begingroup$
Well, I better undelete my original question then!
$endgroup$
– domotorp
8 hours ago
add a comment |
$begingroup$
Well, I better undelete my original question then!
$endgroup$
– domotorp
8 hours ago
$begingroup$
Well, I better undelete my original question then!
$endgroup$
– domotorp
8 hours ago
$begingroup$
Well, I better undelete my original question then!
$endgroup$
– domotorp
8 hours ago
add a comment |
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$begingroup$
If I understand this, you ask for a set with certain properties, one of which is that for every point in the set, the set has "half the points" of the unit circle with the first point as its center. I suspect the answer is that the empty set is such a set, and that there is no other. Gerhard "Of Course Suspicion Isn't Proof" Paseman, 2019.08.17.
$endgroup$
– Gerhard Paseman
9 hours ago