Is the set of paths between any two points moving only in units on the plane countable or uncountable?The set...
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Is the set of paths between any two points moving only in units on the plane countable or uncountable?
The set of points that are equdistiant from two circlesproof of the fact that the set of points equidistant from sides of an angle form a bisector of the angleFinding a point between two points given the points and a distanceWhat is the smaller angle made by the two radial paths?Compute the angle between a line and a plane if the line forms the angles of 45 degrees and 60 degrees with two perpendicular lines lying in the planeProve any line passes through at least two pointsHow many points are possible in the plane that are the same distance from all three points.Existence of a point between two points in Hilbert geometryPoints moving towards the nearest point, where will they meet?Shortest distance between two points , when you can't cross sphere in the middle
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Consider 2 arbitrary, fixed points A and B on the plane. Suppose you can move from point A in unit distance at any angle to another point and from this point you can again travel a unit distance at any angle to another point and so on. Ultimately your goal is to travel from point A to point B along a path of unit-distance line segments without repeating a point during your journey. Is the set of such paths countable or uncountable?
I believe it is uncountable and here is my thought process, but I'm not sure of my logic. Consider the line between the two points equidistant from them; call this line L. A path of only unit distances can be made between A and any point, P, which lies on L without crossing L(I don't know how to prove this statement but it seems true). This path can be mirrored on the other side of L to connect B to P. Thus for any point, P, which lies on L a path can be made from A to B crossing through P halfway through the path. Since the points on L are uncountable the set of paths between A and B are also uncountable.
geometry elementary-set-theory
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$begingroup$
Consider 2 arbitrary, fixed points A and B on the plane. Suppose you can move from point A in unit distance at any angle to another point and from this point you can again travel a unit distance at any angle to another point and so on. Ultimately your goal is to travel from point A to point B along a path of unit-distance line segments without repeating a point during your journey. Is the set of such paths countable or uncountable?
I believe it is uncountable and here is my thought process, but I'm not sure of my logic. Consider the line between the two points equidistant from them; call this line L. A path of only unit distances can be made between A and any point, P, which lies on L without crossing L(I don't know how to prove this statement but it seems true). This path can be mirrored on the other side of L to connect B to P. Thus for any point, P, which lies on L a path can be made from A to B crossing through P halfway through the path. Since the points on L are uncountable the set of paths between A and B are also uncountable.
geometry elementary-set-theory
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add a comment |
$begingroup$
Consider 2 arbitrary, fixed points A and B on the plane. Suppose you can move from point A in unit distance at any angle to another point and from this point you can again travel a unit distance at any angle to another point and so on. Ultimately your goal is to travel from point A to point B along a path of unit-distance line segments without repeating a point during your journey. Is the set of such paths countable or uncountable?
I believe it is uncountable and here is my thought process, but I'm not sure of my logic. Consider the line between the two points equidistant from them; call this line L. A path of only unit distances can be made between A and any point, P, which lies on L without crossing L(I don't know how to prove this statement but it seems true). This path can be mirrored on the other side of L to connect B to P. Thus for any point, P, which lies on L a path can be made from A to B crossing through P halfway through the path. Since the points on L are uncountable the set of paths between A and B are also uncountable.
geometry elementary-set-theory
$endgroup$
Consider 2 arbitrary, fixed points A and B on the plane. Suppose you can move from point A in unit distance at any angle to another point and from this point you can again travel a unit distance at any angle to another point and so on. Ultimately your goal is to travel from point A to point B along a path of unit-distance line segments without repeating a point during your journey. Is the set of such paths countable or uncountable?
I believe it is uncountable and here is my thought process, but I'm not sure of my logic. Consider the line between the two points equidistant from them; call this line L. A path of only unit distances can be made between A and any point, P, which lies on L without crossing L(I don't know how to prove this statement but it seems true). This path can be mirrored on the other side of L to connect B to P. Thus for any point, P, which lies on L a path can be made from A to B crossing through P halfway through the path. Since the points on L are uncountable the set of paths between A and B are also uncountable.
geometry elementary-set-theory
geometry elementary-set-theory
edited 2 hours ago
Andrés E. Caicedo
65.6k8159250
65.6k8159250
asked 2 hours ago
Knight98Knight98
253
253
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5 Answers
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$begingroup$
Yes, this looks convincing. For the missing step, go directly from A towards P in unit steps until the distance left is less than 2. Then use the remaining distance as the base of an isosceles triangle with unit legs, which you make point away from L.
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add a comment |
$begingroup$
Let $C$ be on the line through $B$ that is perpendicular to the segment $AB$ with the distance $BC$ equal to $1/2.$ Take any half-line $L$, not through $B$, that originates at $A$ and intersects the segment $BC.$ For some $nin Bbb N$ there is a path along $L,$ starting at $A,$ determined by $n$ points $A=A_1,...,A_n$ where $A_j,A_{j+1}$ are distance $1$ apart for each $j<n,$ and such that the distance from $A_n$ to $B$ is less than $1.$
Let the point $D$ be such that $A_nD=BD=1$ nd $Dnot in {A_1,...,A_n}.$ Then the path determined by ${A_1,...,A_n}cup {D}$ is a path of the desired type.
The cardinal of the set all such $L$ is $2^{aleph_0}$ so there at least this many paths of the desired type, joining pairs of points .
And each path is determined by a function from some ${1,2,...,m}subset Bbb N$ into $Bbb R^2.$ The set of all such functions has cardinal $2^{aleph_0}$ so there are at most $2^{aleph_0}$ paths of the desired type.
$endgroup$
add a comment |
$begingroup$
Go one unit from a at an angle of t to c.
Go in unit steps along ca until one is less than a unit away from b to a point p.
If p /= b, then draw a triangle with base pb and sides of unit length adding the sides as the final steps.
As for each t in [0,2$pi$), I've constructed a different accepted zigzaging from a to b, there are uncountably many ways of so staggering from a to b.
$endgroup$
add a comment |
$begingroup$
I believe it is uncountable also, and here is my thought process:
For each direction vector $vinBbb R^2=T_aBbb R^2$, take a curve $alpha_v$ from $a$ to $b$ with $alpha_v'(0)=v$.
Then if $vneq w$, we have $alpha_vneqalpha_w$.
But clearly there are uncountably many direction vectors in $Bbb R^2=T_aBbb R^2$.
$endgroup$
$begingroup$
It seems OP wants movements made of straight line segments all of length $1,$ to go from fixed $A$ to fixed $b$ points. [rather than curves as you use]
$endgroup$
– coffeemath
2 hours ago
$begingroup$
I did notice that. But I thought, well, straight lines have derivatives. @coffeemath
$endgroup$
– Chris Custer
1 hour ago
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Cris-- Then to finish one has to show that after the first initial unit straight line step from $A,$ it is possible to continue such steps and somehow arrive at $B.$
$endgroup$
– coffeemath
29 mins ago
$begingroup$
@coffeemath yes. Well there certainly appear to be some details missing. But this seems fairly reasonable. Thanks.
$endgroup$
– Chris Custer
26 mins ago
add a comment |
$begingroup$
Your proof is correct. We can fill in the details on the first part: For any two distinct points $A$ and $B$ in the plane, we can construct a path consisting of unit distances from $A$ to $B$.
If $A$ is more than one unit away from $B$, start at $A$ and take unit steps toward $B$ until we are less than one unit from $B$. If $B$ is a whole number of steps away, we will land on $B$. If not, we will land on a point $P$ within the unit circle centered at $B$. Since $P$ is less than a distance of $1$ away from $B$, the unit circle centered at $P$ will intersect the unit circle centered at $B$ in two places. Take one step from $P$ to an intersection point and then one step from the intersection point to $B$ to finish the path.
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add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, this looks convincing. For the missing step, go directly from A towards P in unit steps until the distance left is less than 2. Then use the remaining distance as the base of an isosceles triangle with unit legs, which you make point away from L.
$endgroup$
add a comment |
$begingroup$
Yes, this looks convincing. For the missing step, go directly from A towards P in unit steps until the distance left is less than 2. Then use the remaining distance as the base of an isosceles triangle with unit legs, which you make point away from L.
$endgroup$
add a comment |
$begingroup$
Yes, this looks convincing. For the missing step, go directly from A towards P in unit steps until the distance left is less than 2. Then use the remaining distance as the base of an isosceles triangle with unit legs, which you make point away from L.
$endgroup$
Yes, this looks convincing. For the missing step, go directly from A towards P in unit steps until the distance left is less than 2. Then use the remaining distance as the base of an isosceles triangle with unit legs, which you make point away from L.
answered 2 hours ago
Henning MakholmHenning Makholm
241k17308547
241k17308547
add a comment |
add a comment |
$begingroup$
Let $C$ be on the line through $B$ that is perpendicular to the segment $AB$ with the distance $BC$ equal to $1/2.$ Take any half-line $L$, not through $B$, that originates at $A$ and intersects the segment $BC.$ For some $nin Bbb N$ there is a path along $L,$ starting at $A,$ determined by $n$ points $A=A_1,...,A_n$ where $A_j,A_{j+1}$ are distance $1$ apart for each $j<n,$ and such that the distance from $A_n$ to $B$ is less than $1.$
Let the point $D$ be such that $A_nD=BD=1$ nd $Dnot in {A_1,...,A_n}.$ Then the path determined by ${A_1,...,A_n}cup {D}$ is a path of the desired type.
The cardinal of the set all such $L$ is $2^{aleph_0}$ so there at least this many paths of the desired type, joining pairs of points .
And each path is determined by a function from some ${1,2,...,m}subset Bbb N$ into $Bbb R^2.$ The set of all such functions has cardinal $2^{aleph_0}$ so there are at most $2^{aleph_0}$ paths of the desired type.
$endgroup$
add a comment |
$begingroup$
Let $C$ be on the line through $B$ that is perpendicular to the segment $AB$ with the distance $BC$ equal to $1/2.$ Take any half-line $L$, not through $B$, that originates at $A$ and intersects the segment $BC.$ For some $nin Bbb N$ there is a path along $L,$ starting at $A,$ determined by $n$ points $A=A_1,...,A_n$ where $A_j,A_{j+1}$ are distance $1$ apart for each $j<n,$ and such that the distance from $A_n$ to $B$ is less than $1.$
Let the point $D$ be such that $A_nD=BD=1$ nd $Dnot in {A_1,...,A_n}.$ Then the path determined by ${A_1,...,A_n}cup {D}$ is a path of the desired type.
The cardinal of the set all such $L$ is $2^{aleph_0}$ so there at least this many paths of the desired type, joining pairs of points .
And each path is determined by a function from some ${1,2,...,m}subset Bbb N$ into $Bbb R^2.$ The set of all such functions has cardinal $2^{aleph_0}$ so there are at most $2^{aleph_0}$ paths of the desired type.
$endgroup$
add a comment |
$begingroup$
Let $C$ be on the line through $B$ that is perpendicular to the segment $AB$ with the distance $BC$ equal to $1/2.$ Take any half-line $L$, not through $B$, that originates at $A$ and intersects the segment $BC.$ For some $nin Bbb N$ there is a path along $L,$ starting at $A,$ determined by $n$ points $A=A_1,...,A_n$ where $A_j,A_{j+1}$ are distance $1$ apart for each $j<n,$ and such that the distance from $A_n$ to $B$ is less than $1.$
Let the point $D$ be such that $A_nD=BD=1$ nd $Dnot in {A_1,...,A_n}.$ Then the path determined by ${A_1,...,A_n}cup {D}$ is a path of the desired type.
The cardinal of the set all such $L$ is $2^{aleph_0}$ so there at least this many paths of the desired type, joining pairs of points .
And each path is determined by a function from some ${1,2,...,m}subset Bbb N$ into $Bbb R^2.$ The set of all such functions has cardinal $2^{aleph_0}$ so there are at most $2^{aleph_0}$ paths of the desired type.
$endgroup$
Let $C$ be on the line through $B$ that is perpendicular to the segment $AB$ with the distance $BC$ equal to $1/2.$ Take any half-line $L$, not through $B$, that originates at $A$ and intersects the segment $BC.$ For some $nin Bbb N$ there is a path along $L,$ starting at $A,$ determined by $n$ points $A=A_1,...,A_n$ where $A_j,A_{j+1}$ are distance $1$ apart for each $j<n,$ and such that the distance from $A_n$ to $B$ is less than $1.$
Let the point $D$ be such that $A_nD=BD=1$ nd $Dnot in {A_1,...,A_n}.$ Then the path determined by ${A_1,...,A_n}cup {D}$ is a path of the desired type.
The cardinal of the set all such $L$ is $2^{aleph_0}$ so there at least this many paths of the desired type, joining pairs of points .
And each path is determined by a function from some ${1,2,...,m}subset Bbb N$ into $Bbb R^2.$ The set of all such functions has cardinal $2^{aleph_0}$ so there are at most $2^{aleph_0}$ paths of the desired type.
edited 2 hours ago
answered 2 hours ago
DanielWainfleetDanielWainfleet
35.3k31648
35.3k31648
add a comment |
add a comment |
$begingroup$
Go one unit from a at an angle of t to c.
Go in unit steps along ca until one is less than a unit away from b to a point p.
If p /= b, then draw a triangle with base pb and sides of unit length adding the sides as the final steps.
As for each t in [0,2$pi$), I've constructed a different accepted zigzaging from a to b, there are uncountably many ways of so staggering from a to b.
$endgroup$
add a comment |
$begingroup$
Go one unit from a at an angle of t to c.
Go in unit steps along ca until one is less than a unit away from b to a point p.
If p /= b, then draw a triangle with base pb and sides of unit length adding the sides as the final steps.
As for each t in [0,2$pi$), I've constructed a different accepted zigzaging from a to b, there are uncountably many ways of so staggering from a to b.
$endgroup$
add a comment |
$begingroup$
Go one unit from a at an angle of t to c.
Go in unit steps along ca until one is less than a unit away from b to a point p.
If p /= b, then draw a triangle with base pb and sides of unit length adding the sides as the final steps.
As for each t in [0,2$pi$), I've constructed a different accepted zigzaging from a to b, there are uncountably many ways of so staggering from a to b.
$endgroup$
Go one unit from a at an angle of t to c.
Go in unit steps along ca until one is less than a unit away from b to a point p.
If p /= b, then draw a triangle with base pb and sides of unit length adding the sides as the final steps.
As for each t in [0,2$pi$), I've constructed a different accepted zigzaging from a to b, there are uncountably many ways of so staggering from a to b.
answered 1 hour ago
William ElliotWilliam Elliot
8,4022720
8,4022720
add a comment |
add a comment |
$begingroup$
I believe it is uncountable also, and here is my thought process:
For each direction vector $vinBbb R^2=T_aBbb R^2$, take a curve $alpha_v$ from $a$ to $b$ with $alpha_v'(0)=v$.
Then if $vneq w$, we have $alpha_vneqalpha_w$.
But clearly there are uncountably many direction vectors in $Bbb R^2=T_aBbb R^2$.
$endgroup$
$begingroup$
It seems OP wants movements made of straight line segments all of length $1,$ to go from fixed $A$ to fixed $b$ points. [rather than curves as you use]
$endgroup$
– coffeemath
2 hours ago
$begingroup$
I did notice that. But I thought, well, straight lines have derivatives. @coffeemath
$endgroup$
– Chris Custer
1 hour ago
$begingroup$
Cris-- Then to finish one has to show that after the first initial unit straight line step from $A,$ it is possible to continue such steps and somehow arrive at $B.$
$endgroup$
– coffeemath
29 mins ago
$begingroup$
@coffeemath yes. Well there certainly appear to be some details missing. But this seems fairly reasonable. Thanks.
$endgroup$
– Chris Custer
26 mins ago
add a comment |
$begingroup$
I believe it is uncountable also, and here is my thought process:
For each direction vector $vinBbb R^2=T_aBbb R^2$, take a curve $alpha_v$ from $a$ to $b$ with $alpha_v'(0)=v$.
Then if $vneq w$, we have $alpha_vneqalpha_w$.
But clearly there are uncountably many direction vectors in $Bbb R^2=T_aBbb R^2$.
$endgroup$
$begingroup$
It seems OP wants movements made of straight line segments all of length $1,$ to go from fixed $A$ to fixed $b$ points. [rather than curves as you use]
$endgroup$
– coffeemath
2 hours ago
$begingroup$
I did notice that. But I thought, well, straight lines have derivatives. @coffeemath
$endgroup$
– Chris Custer
1 hour ago
$begingroup$
Cris-- Then to finish one has to show that after the first initial unit straight line step from $A,$ it is possible to continue such steps and somehow arrive at $B.$
$endgroup$
– coffeemath
29 mins ago
$begingroup$
@coffeemath yes. Well there certainly appear to be some details missing. But this seems fairly reasonable. Thanks.
$endgroup$
– Chris Custer
26 mins ago
add a comment |
$begingroup$
I believe it is uncountable also, and here is my thought process:
For each direction vector $vinBbb R^2=T_aBbb R^2$, take a curve $alpha_v$ from $a$ to $b$ with $alpha_v'(0)=v$.
Then if $vneq w$, we have $alpha_vneqalpha_w$.
But clearly there are uncountably many direction vectors in $Bbb R^2=T_aBbb R^2$.
$endgroup$
I believe it is uncountable also, and here is my thought process:
For each direction vector $vinBbb R^2=T_aBbb R^2$, take a curve $alpha_v$ from $a$ to $b$ with $alpha_v'(0)=v$.
Then if $vneq w$, we have $alpha_vneqalpha_w$.
But clearly there are uncountably many direction vectors in $Bbb R^2=T_aBbb R^2$.
answered 2 hours ago
Chris CusterChris Custer
14.1k3827
14.1k3827
$begingroup$
It seems OP wants movements made of straight line segments all of length $1,$ to go from fixed $A$ to fixed $b$ points. [rather than curves as you use]
$endgroup$
– coffeemath
2 hours ago
$begingroup$
I did notice that. But I thought, well, straight lines have derivatives. @coffeemath
$endgroup$
– Chris Custer
1 hour ago
$begingroup$
Cris-- Then to finish one has to show that after the first initial unit straight line step from $A,$ it is possible to continue such steps and somehow arrive at $B.$
$endgroup$
– coffeemath
29 mins ago
$begingroup$
@coffeemath yes. Well there certainly appear to be some details missing. But this seems fairly reasonable. Thanks.
$endgroup$
– Chris Custer
26 mins ago
add a comment |
$begingroup$
It seems OP wants movements made of straight line segments all of length $1,$ to go from fixed $A$ to fixed $b$ points. [rather than curves as you use]
$endgroup$
– coffeemath
2 hours ago
$begingroup$
I did notice that. But I thought, well, straight lines have derivatives. @coffeemath
$endgroup$
– Chris Custer
1 hour ago
$begingroup$
Cris-- Then to finish one has to show that after the first initial unit straight line step from $A,$ it is possible to continue such steps and somehow arrive at $B.$
$endgroup$
– coffeemath
29 mins ago
$begingroup$
@coffeemath yes. Well there certainly appear to be some details missing. But this seems fairly reasonable. Thanks.
$endgroup$
– Chris Custer
26 mins ago
$begingroup$
It seems OP wants movements made of straight line segments all of length $1,$ to go from fixed $A$ to fixed $b$ points. [rather than curves as you use]
$endgroup$
– coffeemath
2 hours ago
$begingroup$
It seems OP wants movements made of straight line segments all of length $1,$ to go from fixed $A$ to fixed $b$ points. [rather than curves as you use]
$endgroup$
– coffeemath
2 hours ago
$begingroup$
I did notice that. But I thought, well, straight lines have derivatives. @coffeemath
$endgroup$
– Chris Custer
1 hour ago
$begingroup$
I did notice that. But I thought, well, straight lines have derivatives. @coffeemath
$endgroup$
– Chris Custer
1 hour ago
$begingroup$
Cris-- Then to finish one has to show that after the first initial unit straight line step from $A,$ it is possible to continue such steps and somehow arrive at $B.$
$endgroup$
– coffeemath
29 mins ago
$begingroup$
Cris-- Then to finish one has to show that after the first initial unit straight line step from $A,$ it is possible to continue such steps and somehow arrive at $B.$
$endgroup$
– coffeemath
29 mins ago
$begingroup$
@coffeemath yes. Well there certainly appear to be some details missing. But this seems fairly reasonable. Thanks.
$endgroup$
– Chris Custer
26 mins ago
$begingroup$
@coffeemath yes. Well there certainly appear to be some details missing. But this seems fairly reasonable. Thanks.
$endgroup$
– Chris Custer
26 mins ago
add a comment |
$begingroup$
Your proof is correct. We can fill in the details on the first part: For any two distinct points $A$ and $B$ in the plane, we can construct a path consisting of unit distances from $A$ to $B$.
If $A$ is more than one unit away from $B$, start at $A$ and take unit steps toward $B$ until we are less than one unit from $B$. If $B$ is a whole number of steps away, we will land on $B$. If not, we will land on a point $P$ within the unit circle centered at $B$. Since $P$ is less than a distance of $1$ away from $B$, the unit circle centered at $P$ will intersect the unit circle centered at $B$ in two places. Take one step from $P$ to an intersection point and then one step from the intersection point to $B$ to finish the path.
$endgroup$
add a comment |
$begingroup$
Your proof is correct. We can fill in the details on the first part: For any two distinct points $A$ and $B$ in the plane, we can construct a path consisting of unit distances from $A$ to $B$.
If $A$ is more than one unit away from $B$, start at $A$ and take unit steps toward $B$ until we are less than one unit from $B$. If $B$ is a whole number of steps away, we will land on $B$. If not, we will land on a point $P$ within the unit circle centered at $B$. Since $P$ is less than a distance of $1$ away from $B$, the unit circle centered at $P$ will intersect the unit circle centered at $B$ in two places. Take one step from $P$ to an intersection point and then one step from the intersection point to $B$ to finish the path.
$endgroup$
add a comment |
$begingroup$
Your proof is correct. We can fill in the details on the first part: For any two distinct points $A$ and $B$ in the plane, we can construct a path consisting of unit distances from $A$ to $B$.
If $A$ is more than one unit away from $B$, start at $A$ and take unit steps toward $B$ until we are less than one unit from $B$. If $B$ is a whole number of steps away, we will land on $B$. If not, we will land on a point $P$ within the unit circle centered at $B$. Since $P$ is less than a distance of $1$ away from $B$, the unit circle centered at $P$ will intersect the unit circle centered at $B$ in two places. Take one step from $P$ to an intersection point and then one step from the intersection point to $B$ to finish the path.
$endgroup$
Your proof is correct. We can fill in the details on the first part: For any two distinct points $A$ and $B$ in the plane, we can construct a path consisting of unit distances from $A$ to $B$.
If $A$ is more than one unit away from $B$, start at $A$ and take unit steps toward $B$ until we are less than one unit from $B$. If $B$ is a whole number of steps away, we will land on $B$. If not, we will land on a point $P$ within the unit circle centered at $B$. Since $P$ is less than a distance of $1$ away from $B$, the unit circle centered at $P$ will intersect the unit circle centered at $B$ in two places. Take one step from $P$ to an intersection point and then one step from the intersection point to $B$ to finish the path.
answered 2 hours ago
John DoumaJohn Douma
5,56211319
5,56211319
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